IB Analysis and Approaches HL · Geometry in space · End of topic

IB exam questions: topic 8

146 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

Level

Paper

Solutions

Vectors: components, magnitude and operations

Exercise 1 BasicIB · P1 TZ2 May 2010

The three vectors $a$ , $b$ and $c$ are given by

$a = 2 y - 3 x 2 x$ , $b = 4 x y 3 - x$ and $c = 4 - 7 6$ , where $x, y \in R$ .

a) If $a + 2 b - c = 0$ , find the value of $x$ and of $y$ .

b) Find the exact value of $∣ a + 2 b ∣$ .

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Answer: a) $x = 1$ , $y = - 2$ ; b) $101$

Show working and answer

a) 1. We calculate $a + 2 b - c$ component by component and equate it to $0$ : $2 y + 8 x - 4 - 3 x + 2 y + 7 2 x + 2 (3 - x) - 6 = 000 .$ 2. The third component is $2 x + 6 - 2 x - 6 = 0$ , which always holds. Two equations remain: ${8 x + 2 y - 3 x + 2 y = 4 = - 7$ 3. Subtracting the second from the first: $11 x = 11$ , so $x = 1$ . 4. We substitute into $8 x + 2 y = 4$ : $8 + 2 y = 4$ , hence $y = - 2$ .

b) 1. As $a + 2 b - c = 0$ , we have $a + 2 b = c$ . 2. Therefore $∣ a + 2 b ∣ = ∣ c ∣ = 4^{2} + (- 7)^{2} + 6^{2} = 16 + 49 + 36 = 101$ .

Answer: a) $x = 1$ , $y = - 2$ ; b) $101$

Exercise 2 HardIB · P1 TZ0 November 2015

Consider the triangle $ABC$ . The points $P$ , $Q$ and $R$ are the midpoints of the line segments $[AB]$ , $[BC]$ and $[AC]$ respectively.

Let $OA = a$ , $OB = b$ and $OC = c$ .

a) Find $BR$ in terms of $a$ , $b$ and $c$ .

b) (i) Find a vector equation of the line passing through $B$ and $R$ in terms of $a$ , $b$ , $c$ and a parameter $λ$ .

(ii) Find a vector equation of the line passing through $A$ and $Q$ in terms of $a$ , $b$ , $c$ and a parameter $μ$ .

(iii) Hence, show that $OG = \frac{1}{3} (a + b + c)$ , given that $G$ is the point where $[BR]$ and $[AQ]$ intersect.

c) Show that the line segment $[CP]$ also contains the point $G$ .

The coordinates of the points $A$ , $B$ and $C$ are $(1, 3, 1)$ , $(3, 7, - 5)$ and $(2, 2, 1)$ respectively. A point $X$ is such that $[GX]$ is perpendicular to the plane $ABC$ .

d) Given that the tetrahedron $ABCX$ has a volume of $12$ units $^{3}$ , find the possible coordinates of $X$ .

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Answer: a) $B R = \frac{1}{2} a - b + \frac{1}{2} c$ ; b) (i) $r = \frac{λ}{2} a + (1 - λ) b + \frac{λ}{2} c$ ; (ii) $r = (1 - μ) a + \frac{μ}{2} b + \frac{μ}{2} c$ ; d) $X (6, 8, 3)$ or $X (- 2, 0, - 5)$ . Parts b) (iii) and c) are proved.

Show working and answer

a) 1. $R$ is the midpoint of $[A C]$ : $O R = \frac{1}{2} (a + c)$ . 2. $B R = O R - O B = \frac{1}{2} a - b + \frac{1}{2} c$ .

b) (i) The line passes through $B$ with direction $B R$ : $r = b + λ (\frac{1}{2} a - b + \frac{1}{2} c) = \frac{λ}{2} a + (1 - λ) b + \frac{λ}{2} c .$

(ii) 1. $Q$ is the midpoint of $[B C]$ : $O Q = \frac{1}{2} (b + c)$ , and $A Q = \frac{1}{2} b + \frac{1}{2} c - a$ . 2. $r = a + μ A Q = (1 - μ) a + \frac{μ}{2} b + \frac{μ}{2} c$ .

(iii) 1. At the point $G$ the two expressions coincide. Equating the coefficients of $a$ , $b$ and $c$ : $\frac{λ}{2} = 1 - μ, 1 - λ = \frac{μ}{2}, \frac{λ}{2} = \frac{μ}{2} .$ 2. From the third, $λ = μ$ . In the first: $\frac{λ}{2} = 1 - λ$ , so $λ = \frac{2}{3}$ . The second also holds: $1 - \frac{2}{3} = \frac{1}{3} = \frac{2/3}{2}$ . 3. With $λ = \frac{2}{3}$ : $O G = \frac{1}{3} a + \frac{1}{3} b + \frac{1}{3} c = \frac{1}{3} (a + b + c)$ .

c) 1. $P$ is the midpoint of $[A B]$ : $O P = \frac{1}{2} (a + b)$ , so $C P = \frac{1}{2} a + \frac{1}{2} b - c$ . 2. $C G = O G - O C = \frac{1}{3} (a + b + c) - c = \frac{1}{3} (a + b - 2 c)$ . 3. Then $C P = \frac{1}{2} (a + b - 2 c)$ , and therefore $C G = \frac{2}{3} C P$ . As $0 < \frac{2}{3} < 1$ , the point $G$ lies on the segment $[C P]$ .

d) 1. $G = \frac{1}{3} (A + B + C) = \frac{1}{3} (6, 12, - 3) = (2, 4, - 1)$ . 2. $A B = (2, 4, - 6)$ , $A C = (1, - 1, 0)$ and $A B \times A C = 4 \cdot 0 - (- 6) (- 1) (- 6) \cdot 1 - 2 \cdot 0 2 \cdot (- 1) - 4 \cdot 1 = - 6 - 6 - 6$ . 3. This direction is $(1, 1, 1)$ , normal to the plane $A B C$ , and the area of the triangle is $\frac{1}{2} 108 = 33$ . 4. Volume of the tetrahedron: $\frac{1}{3} \cdot 33 \cdot h = 12$ , so $h = \frac{12}{3} = 43$ , and $h = ∣ GX ∣$ . 5. $X = G \pm 43 \cdot \frac{( 1 , 1 , 1 )}{3} = (2, 4, - 1) \pm (4, 4, 4)$ . 6. $X = (6, 8, 3)$ or $X = (- 2, 0, - 5)$ .

Scalar product, angles and perpendicularity

Exercise 3 BasicIB · P1 May 2002

Find the angle between the vectors $v = i + j + 2 k$ and $w = 2 i + 3 j + k$ . Give your answer in radians.

Show answer

Answer: $θ \approx 0.702$ rad

Show working and answer

Scalar product: $v \cdot w = 2 + 3 + 2 = 7$ .
Magnitudes: $∣ v ∣ = 1 + 1 + 4 = 6$ and $∣ w ∣ = 4 + 9 + 1 = 14$ .
$cos θ = \frac{7}{6 14} = \frac{7}{84} = \frac{21}{6} \approx 0.764$ .
With the calculator in radian mode: $θ = arccos \frac{21}{6} \approx 0.702$ rad.

Answer: $θ \approx 0.702$ rad

Exercise 4 BasicIB · P1 TZ1 May 2019

Let $a = 2 k - 1$ and $b = - 3 k + 2 k$ , $k \in R$ .

Given that $a$ and $b$ are perpendicular, find the possible values of $k$ .

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Answer: $k = 2$ and $k = - 3$

Show working and answer

Perpendicular: $a \cdot b = 0$ , i.e. $- 6 + k (k + 2) - k = 0$ .
Simplify: $k^{2} + k - 6 = 0$ .
Factorise: $(k - 2) (k + 3) = 0$ .

Answer: $k = 2$ and $k = - 3$

Exercise 5 BasicIB · P1 TZ2 May 2018

The acute angle between the vectors $3 i - 4 j - 5 k$ and $5 i - 4 j + 3 k$ is denoted $θ$ . Find $cos θ$ .

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Answer: $cos θ = \frac{8}{25}$

Show working and answer

Scalar product: $15 + 16 - 15 = 16$ .
Magnitudes: $9 + 16 + 25 = 50$ and $25 + 16 + 9 = 50$ .
$cos θ = \frac{16}{50 50} = \frac{16}{50} = \frac{8}{25}$ . It is positive, as expected for an acute angle.

Answer: $cos θ = \frac{8}{25}$

Exercise 6 BasicIB · P1 TZ1 May 2010Proof

Consider the vectors $O A = a$ , $O B = b$ and $O C = a + b$ . Show that if $∣ a ∣ = ∣ b ∣$ then $(a + b) \cdot (a - b) = 0$ . Comment on what this tells us about the parallelogram $OACB$ .

View solution

We expand the scalar product (it is distributive and commutative): $(a + b) \cdot (a - b) = a \cdot a - a \cdot b + b \cdot a - b \cdot b = ∣ a ∣^{2} - ∣ b ∣^{2} .$
If $∣ a ∣ = ∣ b ∣$ , then $∣ a ∣^{2} - ∣ b ∣^{2} = 0$ , so $(a + b) \cdot (a - b) = 0$ .
Interpretation: $a + b = O C$ is one diagonal of the parallelogram $OACB$ , and $a - b = B A$ is the other. Their scalar product is zero, so the two diagonals are perpendicular. Moreover, $∣ a ∣ = ∣ b ∣$ means that two adjacent sides are equal, that is, the parallelogram is a rhombus (and the diagonals of a rhombus meet at right angles).

As required.

Exercise 7 MediumIB · P2 TZ0 November 2008

The angle between the vector $a = i - 2 j + 3 k$ and the vector $b = 3 i - 2 j + m k$ is $3 0^{\circ}$ . Find the values of $m$ .

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Answer: $m \approx 25.7$ and $m \approx 2.27$

Show working and answer

We use $cos θ = \frac{a \cdot b}{∣ a ∣ ∣ b ∣}$ with $θ = 3 0^{\circ}$ .
We calculate: $a \cdot b = 3 + 4 + 3 m = 7 + 3 m$ , $∣ a ∣ = 1 + 4 + 9 = 14$ and $∣ b ∣ = 13 + m^{2}$ .
We set up $\frac{7 + 3 m}{14 13 + m ^{2}} = cos 3 0^{\circ} = \frac{3}{2} .$
We square both sides (we will check afterwards that $7 + 3 m > 0$ ): $4 (7 + 3 m)^{2} = 3 \cdot 14 (13 + m^{2})$ , i.e. $196 + 168 m + 36 m^{2} = 546 + 42 m^{2}$ .
Rearranging: $6 m^{2} - 168 m + 350 = 0$ , or $3 m^{2} - 84 m + 175 = 0$ .
We solve (by hand using the formula or with the GDC): $m = \frac{84 \pm 4956}{6} = 14 \pm \frac{1239}{3}$ , i.e. $m \approx 25.7$ and $m \approx 2.27$ .
Check: for both values $7 + 3 m > 0$ , so the angle is acute and both solutions are valid.

Answer: $m \approx 25.7$ and $m \approx 2.27$

Exercise 8 MediumIB · P1 TZ2 May 2005

The position vectors of the points $P$ and $Q$ are $2 - 3 1$ and $22 - 4$ respectively. The origin is at $O$ .

Find:

a) the angle $POQ$ .

b) the area of triangle $OPQ$ .

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Answer: a) $POQ \approx 109. 1^{\circ}$ ; b) $53 \approx 8.66$

Show working and answer

a) 1. The vectors are $O P = 2 - 3 1$ and $O Q = 22 - 4$ . 2. Scalar product: $O P \cdot O Q = 4 - 6 - 4 = - 6$ . 3. Magnitudes: $∣ O P ∣ = 4 + 9 + 1 = 14$ and $∣ O Q ∣ = 4 + 4 + 16 = 24$ . 4. $cos P O Q = \frac{- 6}{14 24} = - \frac{21}{14} \approx - 0.327$ , and therefore $P O Q = arccos (- \frac{21}{14}) \approx 109. 1^{\circ}$ .

b) 1. The area of the triangle is half the magnitude of the vector product: $O P \times O Q = (- 3) (- 4) - 1 \cdot 2 1 \cdot 2 - 2 \cdot (- 4) 2 \cdot 2 - (- 3) \cdot 2 = 101010 .$ 2. $∣ O P \times O Q ∣ = 103$ , so the area is $\frac{1}{2} \cdot 103 = 53 \approx 8.66$ .

Answer: a) $POQ \approx 109. 1^{\circ}$ ; b) $53 \approx 8.66$

Exercise 9 MediumIB · P1 TZ0 November 2010

Let $α$ be the angle between the unit vectors $a$ and $b$ , where $0 \leq α \leq π$ .

a) Express $∣ a - b ∣$ and $∣ a + b ∣$ in terms of $α$ .

b) Hence, determine the values of $cos α$ for which $∣ a + b ∣ = 3 ∣ a - b ∣$ .

Show answer

Answer: a) $∣ a - b ∣ = 2 - 2 cos α$ and $∣ a + b ∣ = 2 + 2 cos α$ ; b) $cos α = \frac{4}{5}$

Show working and answer

a) 1. As $a$ and $b$ are unit vectors, $∣ a ∣ = ∣ b ∣ = 1$ and $a \cdot b = cos α$ . 2. We expand the square of the magnitude using the scalar product: $∣ a - b ∣^{2} = ∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2} = 2 - 2 cos α, ∣ a + b ∣^{2} = 2 + 2 cos α .$ 3. As $0 \leq α \leq π$ , both quantities are non-negative, and $∣ a - b ∣ = 2 - 2 cos α, ∣ a + b ∣ = 2 + 2 cos α .$

b) 1. We square the condition $∣ a + b ∣ = 3 ∣ a - b ∣$ : $∣ a + b ∣^{2} = 9 ∣ a - b ∣^{2}$ . 2. We substitute: $2 + 2 cos α = 9 (2 - 2 cos α) = 18 - 18 cos α$ . 3. Grouping terms: $20 cos α = 16$ , so $cos α = \frac{4}{5}$ .

Answer: a) $∣ a - b ∣ = 2 - 2 cos α$ and $∣ a + b ∣ = 2 + 2 cos α$ ; b) $cos α = \frac{4}{5}$

Exercise 10 MediumIB · P2 TZ1 May 2015

Let $v = 235$ and $w = 4 λ 10$ .

a) Find the value of $λ$ for which $v$ and $w$ are parallel.

b) Calculate the value of $λ$ for which $v$ and $w$ are perpendicular.

c) Find the two values of $λ$ if the angle between $v$ and $w$ is $1 0^{\circ}$ .

Show answer

Answer: a) $λ = 6$ ; b) $λ = - \frac{58}{3}$ ; c) $λ \approx 3.73$ and $λ \approx 8.76$

Show working and answer

a) 1. Parallel means $w = k v$ . As $4 = 2 k$ , $k = 2$ (and the third component confirms it: $10 = 2 \cdot 5$ ). 2. Therefore $λ = 2 \cdot 3 = 6$ .

b) 1. Perpendicular means $v \cdot w = 0$ : $8 + 3 λ + 50 = 0$ . 2. Hence $3 λ = - 58$ and $λ = - \frac{58}{3}$ .

c) 1. We use $cos 1 0^{\circ} = \frac{v \cdot w}{∣ v ∣ ∣ w ∣}$ , with $v \cdot w = 58 + 3 λ$ , $∣ v ∣ = 38$ and $∣ w ∣ = 116 + λ^{2}$ : $\frac{58 + 3 λ}{38 116 + λ ^{2}} = cos 1 0^{\circ} .$ 2. This equation is solved with the GDC (for example, by finding the intersections of the graph of the left-hand side with the line $y = cos 1 0^{\circ}$ ), or by squaring, which gives a quadratic equation in $λ$ : $(58 + 3 λ)^{2} = 38 cos^{2} 1 0^{\circ} (116 + λ^{2}) .$ 3. The solutions are $λ \approx 3.73$ and $λ \approx 8.76$ . 4. Check: in both cases $58 + 3 λ > 0$ , so the angle is acute and both solutions are valid.

Answer: a) $λ = 6$ ; b) $λ = - \frac{58}{3}$ ; c) $λ \approx 3.73$ and $λ \approx 8.76$

Exercise 11 MediumIB · P1 TZ2 May 2012

a) Let $a = 13 k$ , $b = 20 - 1$ and $c = - 1 23$ . If $b$ is perpendicular to $c - 2 a$ , calculate the value of $k$ .

b) Find the values of $x$ for which the vectors $1 2 cos x 0$ and $- 1 2 sin x 1$ , $0 \leq x \leq \frac{π}{2}$ are perpendicular.

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Answer: a) $k = \frac{9}{2}$ ; b) $x = \frac{π}{12}$ and $x = \frac{5 π}{12}$

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a) 1. $c - 2 a = - 1 - 2 2 - 6 3 - 2 k = - 3 - 4 3 - 2 k$ . 2. Perpendicular: $b \cdot (c - 2 a) = 0$ , i.e. $- 6 + 0 - (3 - 2 k) = 0$ . 3. Hence $- 9 + 2 k = 0$ and $k = \frac{9}{2}$ .

b) 1. The scalar product must be zero: $- 1 + 4 sin x cos x = 0$ . 2. As $2 sin x cos x = sin 2 x$ , this gives $- 1 + 2 sin 2 x = 0$ , so $sin 2 x = \frac{1}{2}$ . 3. If $0 \leq x \leq \frac{π}{2}$ , then $0 \leq 2 x \leq π$ , and in that interval $sin 2 x = \frac{1}{2}$ for $2 x = \frac{π}{6}$ or $2 x = \frac{5 π}{6}$ . 4. Therefore $x = \frac{π}{12}$ or $x = \frac{5 π}{12}$ .

Answer: a) $k = \frac{9}{2}$ ; b) $x = \frac{π}{12}$ and $x = \frac{5 π}{12}$

Exercise 12 MediumIB · P1 TZ0 November 2018

The vectors $a$ and $b$ are defined by $a = 11 t$ , $b = 0 - t 4 t$ , where $t \in R$ .

a) Find $a \cdot b$ in terms of $t$ and then simplify this expression.

b) Hence, or otherwise, find all the values of $t$ for which the angle between $a$ and $b$ is obtuse.

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Answer: a) $a \cdot b = 4 t^{2} - t$ ; b) $0 < t < \frac{1}{4}$

Show working and answer

a) 1. $a \cdot b = 1 \cdot 0 + 1 \cdot (- t) + t \cdot 4 t = - t + 4 t^{2}$ .

b) 1. The angle is obtuse if $cos θ < 0$ , that is, if $a \cdot b < 0$ (with both vectors non-zero). 2. We solve $4 t^{2} - t < 0$ , that is, $t (4 t - 1) < 0$ . The parabola $y = 4 t^{2} - t$ crosses the axis at $t = 0$ and $t = \frac{1}{4}$ , and is negative between the two roots. 3. So $0 < t < \frac{1}{4}$ . For these values $b \neq = 0$ (and $a$ is never zero), so they are valid.

Answer: a) $a \cdot b = 4 t^{2} - t$ ; b) $0 < t < \frac{1}{4}$

Exercise 13 MediumIB · P1 November 1998

Given the vectors $a = - 2 i + p j - k$ and $b = i + 3 (p + 4) j + (2 p - 5) k$ , calculate the values of $p$ for which $a$ is perpendicular to $b$ .

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Answer: $p = - \frac{1}{3}$ and $p = - 3$

Show working and answer

Perpendicular vectors: $a \cdot b = 0$ : $- 2 \cdot 1 + p \cdot 3 (p + 4) + (- 1) (2 p - 5) = 0.$
We expand: $- 2 + 3 p^{2} + 12 p - 2 p + 5 = 0$ , that is, $3 p^{2} + 10 p + 3 = 0$ .
Solving: $p = \frac{- 10 \pm 100 - 36}{6} = \frac{- 10 \pm 8}{6}$ , so $p = - \frac{1}{3}$ or $p = - 3$ .

Answer: $p = - \frac{1}{3}$ and $p = - 3$

Exercise 14 MediumIB · P1 TZ1 May 2005

The vectors $a$ , $b$ and $c$ are defined by $a = 32 - 1$ , $b = 152$ and $c = 2 y 3$ .

Given that $c$ is perpendicular to $2 a - b$ , find the value of $y$ .

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Answer: $y = - 2$

Show working and answer

$2 a - b = 6 - 1 4 - 5 - 2 - 2 = 5 - 1 - 4$ .
Perpendicular vectors: $c \cdot (2 a - b) = 0$ , that is, $10 - y - 12 = 0$ .
So $y = - 2$ .

Answer: $y = - 2$

Exercise 15 MediumIB · P1 TZ1 May 2007

Consider the vectors $a$ , $b$ , $c$ , $d$

$a = 23 - 1, b = 125, c = 31 λ, d = μ - 2 1$

Let $s = (a \cdot b) c + d$ , where $s$ is perpendicular to $a$ .

Find an expression for $λ$ in terms of $μ$ .

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Answer: $λ = \frac{20 + 2 μ}{3}$

Show working and answer

We calculate the scalar $a \cdot b = 2 \cdot 1 + 3 \cdot 2 + (- 1) \cdot 5 = 2 + 6 - 5 = 3$ .
Then $s = 3 31 λ + μ - 2 1 = 9 + μ 1 3 λ + 1 .$
Since $s ⊥ a$ , their scalar product is zero: $2 (9 + μ) + 3 \cdot 1 - (3 λ + 1) = 0$ .
Simplifying: $18 + 2 μ + 3 - 3 λ - 1 = 0$ , that is, $20 + 2 μ - 3 λ = 0$ .
We solve for $λ$ .

Answer: $λ = \frac{20 + 2 μ}{3}$

Exercise 16 HardIB · P2 TZ0 November 2010

Consider the vectors $a = sin (2 α) i - cos (2 α) j + k$ and $b = cos α i - sin α j - k$ , where $0 < α < 2 π$ .

Let $θ$ be the angle between the vectors $a$ and $b$ .

a) Express $cos θ$ in terms of $α$ .

b) Find the acute angle $α$ for which the two vectors are perpendicular.

c) For $α = \frac{7 π}{6}$ , determine the vector product of $a$ and $b$ .

Show answer

Answer: a) $cos θ = \frac{sin 3 α - 1}{2}$ ; b) $α = \frac{π}{6}$ ; c) $a \times b = 000$

Show working and answer

a) 1. Scalar product: $a \cdot b = sin 2 α cos α + cos 2 α sin α - 1 = sin (2 α + α) - 1 = sin 3 α - 1.$ 2. Magnitudes: $∣ a ∣^{2} = sin^{2} 2 α + cos^{2} 2 α + 1 = 2$ and $∣ b ∣^{2} = cos^{2} α + sin^{2} α + 1 = 2$ . Therefore $∣ a ∣ ∣ b ∣ = 2$ . 3. Hence $cos θ = \frac{a \cdot b}{∣ a ∣ ∣ b ∣} = \frac{sin 3 α - 1}{2}$ .

b) 1. Perpendicular vectors means $cos θ = 0$ , i.e. $sin 3 α = 1$ . 2. If $α$ is acute, $0 < α < \frac{π}{2}$ , then $0 < 3 α < \frac{3 π}{2}$ , and in that interval $sin 3 α = 1$ only for $3 α = \frac{π}{2}$ . 3. Hence $α = \frac{π}{6}$ .

c) 1. For $α = \frac{7 π}{6}$ : $2 α = \frac{7 π}{3}$ , with $sin \frac{7 π}{3} = \frac{3}{2}$ and $cos \frac{7 π}{3} = \frac{1}{2}$ ; also $cos \frac{7 π}{6} = - \frac{3}{2}$ and $sin \frac{7 π}{6} = - \frac{1}{2}$ . 2. Therefore $a = \frac{3}{2} - \frac{1}{2} 1$ and $b = - \frac{3}{2} \frac{1}{2} - 1 = - a$ . 3. (Also: $sin \frac{7 π}{2} = - 1$ , so $cos θ = - 1$ and the vectors are opposite.) Two parallel vectors have a zero vector product: $a \times b = a \times (- a) = 000 .$

Answer: a) $cos θ = \frac{sin 3 α - 1}{2}$ ; b) $α = \frac{π}{6}$ ; c) $a \times b = 000$

Vector product, areas and volumes

Exercise 17 BasicIB · P1 May 1999

a) Calculate a vector perpendicular to both of the vectors $O P = i - 3 j + 2 k$ and $O Q = - 2 i + j - k$ .

b) If $O P$ and $O Q$ are position vectors for the points $P$ and $Q$ , use your answer to part (a) or otherwise, to find the area of triangle $OPQ$ .

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Answer: a) $O P \times O Q = 1 - 3 - 5$ (or any multiple of it); b) area $= \frac{35}{2}$

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a) 1. The vector product is perpendicular to both vectors: $O P \times O Q = 1 - 3 2 \times - 2 1 - 1 = (- 3) (- 1) - 2 \cdot 1 2 \cdot (- 2) - 1 \cdot (- 1) 1 \cdot 1 - (- 3) (- 2) = 1 - 3 - 5 .$

b) 1. The area of the triangle is $\frac{1}{2} ∣ O P \times O Q ∣$ . 2. $∣ O P \times O Q ∣ = 1 + 9 + 25 = 35$ , so the area is $\frac{35}{2}$ .

Answer: a) $O P \times O Q = 1 - 3 - 5$ (or any multiple of it); b) area $= \frac{35}{2}$

Exercise 18 BasicIB · P1 November 2003

Consider the points $A (1, 2, - 4)$ , $B (1, 5, 0)$ and $C (6, 5, - 12)$ . Find the area of $△ ABC$ .

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Answer: area $△ ABC = \frac{1921}{2} \approx 21.9$

Show working and answer

We calculate two sides from $A$ : $A B = 034$ and $A C = 53 - 8$ .
Vector product: $A B \times A C = 3 (- 8) - 4 \cdot 3 4 \cdot 5 - 0 \cdot (- 8) 0 \cdot 3 - 3 \cdot 5 = - 36 20 - 15 .$
Its magnitude is $1296 + 400 + 225 = 1921$ .
The area of the triangle is half of this: $\frac{1921}{2} \approx 21.9$ .

Answer: area $△ ABC = \frac{1921}{2} \approx 21.9$

Exercise 19 BasicIB · P1 May 2003

Given $a = i + 2 j - k$ , $b = - 3 i + 2 j + 2 k$ and $c = 2 i - 3 j + 4 k$ , find $(a \times b) \cdot c$ .

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Answer: $(a \times b) \cdot c = 41$

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Vector product: $a \times b = 2 \cdot 2 - (- 1) \cdot 2 (- 1) (- 3) - 1 \cdot 2 1 \cdot 2 - 2 \cdot (- 3) = 618 .$
Scalar product with $c = 2 - 3 4$ : $(a \times b) \cdot c = 12 - 3 + 32 = 41$ .

Answer: $(a \times b) \cdot c = 41$

Exercise 20 BasicIB · P1 TZ1 May 2004

Given that $a = i + 2 j + k$ , $b = i - 3 j + 2 k$ and $c = 2 i + j - 2 k$ , calculate $(a - b) \cdot (b \times c)$ .

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Answer: $(a - b) \cdot (b \times c) = 23$

Show working and answer

Difference: $a - b = 05 - 1$ .
Vector product: $b \times c = (- 3) (- 2) - 2 \cdot 1 2 \cdot 2 - 1 \cdot (- 2) 1 \cdot 1 - (- 3) \cdot 2 = 467 .$
Scalar product: $(a - b) \cdot (b \times c) = 0 \cdot 4 + 5 \cdot 6 + (- 1) \cdot 7 = 30 - 7 = 23$ .

Answer: $(a - b) \cdot (b \times c) = 23$

Exercise 21 BasicIB · P1 TZ1 May 2013

Consider the points $A (1, 2, 3)$ , $B (1, 0, 5)$ and $C (2, - 1, 4)$ .

a) Calculate $A B \times A C$ .

b) Using part (a), find the area of triangle $ABC$ .

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Answer: a) $A B \times A C = 4 i + 2 j + 2 k$ ; b) area $= 6$

Show working and answer

a) 1. $A B = 0 - 2 2$ and $A C = 1 - 3 1$ . 2. Vector product: $A B \times A C = (- 2) \cdot 1 - 2 \cdot (- 3) 2 \cdot 1 - 0 \cdot 1 0 \cdot (- 3) - (- 2) \cdot 1 = 422 .$

b) 1. Area $= \frac{1}{2} ∣ A B \times A C ∣ = \frac{1}{2} 16 + 4 + 4 = \frac{1}{2} 24 = 6$ .

Answer: a) $A B \times A C = 4 i + 2 j + 2 k$ ; b) area $= 6$

Exercise 22 BasicIB · P1 TZ2 May 2019

Three points in three-dimensional space have coordinates $A (0, 0, 2)$ , $B (0, 2, 0)$ and $C (3, 1, 0)$ .

a) Find the vector (i) $A B$ ; (ii) $A C$ .

b) Hence, or otherwise, find the area of triangle $ABC$ .

Show answer

Answer: a) $A B = 02 - 2$ , $A C = 31 - 2$ ; b) area $= 19$

Show working and answer

a) 1. $A B = B - A = 02 - 2$ . 2. $A C = C - A = 31 - 2$ .

b) 1. Vector product: $A B \times A C = 2 \cdot (- 2) - (- 2) \cdot 1 (- 2) \cdot 3 - 0 \cdot (- 2) 0 \cdot 1 - 2 \cdot 3 = - 2 - 6 - 6 .$ 2. Its magnitude: $4 + 36 + 36 = 76 = 219$ . 3. Area $= \frac{1}{2} \cdot 219 = 19$ .

Answer: a) $A B = 02 - 2$ , $A C = 31 - 2$ ; b) area $= 19$

Exercise 23 BasicIB · P1 May 1996

Calculate the area of the parallelogram determined by the vectors $a = 3 i - j + 2 k$ and $b = 2 i + j - 4 k$ .

Show answer

Answer: area $= 285 \approx 16.9$

Show working and answer

The area of the parallelogram is $∣ a \times b ∣$ .
Cross product: $a \times b = (- 1) (- 4) - 2 \cdot 1 2 \cdot 2 - 3 \cdot (- 4) 3 \cdot 1 - (- 1) \cdot 2 = 2165 .$
Magnitude: $4 + 256 + 25 = 285 \approx 16.9$ .

Answer: area $= 285 \approx 16.9$

Exercise 24 BasicIB · P2 TZ1 May 2016

Points $A$ and $B$ have position vectors $O A = 12 - 2$ and $O B = 102$ .

a) Calculate $O A \times O B$ .

b) Hence find the area of triangle $OAB$ .

Show answer

Answer: a) $O A \times O B = 4 i - 4 j - 2 k$ ; b) area $= 3$

Show working and answer

a) 1. Cross product: $O A \times O B = 2 \cdot 2 - (- 2) \cdot 0 (- 2) \cdot 1 - 1 \cdot 2 1 \cdot 0 - 2 \cdot 1 = 4 - 4 - 2 .$

b) 1. Area $= \frac{1}{2} ∣ O A \times O B ∣ = \frac{1}{2} 16 + 16 + 4 = \frac{1}{2} \cdot 6 = 3$ .

Answer: a) $O A \times O B = 4 i - 4 j - 2 k$ ; b) area $= 3$

Exercise 25 BasicIB · P1 May 1997

The coordinates of the points $P$ and $Q$ are $(- 1, 5, 7)$ and $(1, 2, 3)$ respectively, and $O$ is the origin.

a) Calculate the vectors $O P$ and $O Q$ .

b) Calculate the area of the parallelogram that has adjacent sides $O P$ and $O Q$ .

Show answer

Answer: a) $O P = - 1 57$ , $O Q = 123$ ; b) area $= 56$

Show working and answer

a) 1. Since $O$ is the origin, the vectors coincide with the coordinates: $O P = - 1 57$ and $O Q = 123$ .

b) 1. The area of the parallelogram is $∣ O P \times O Q ∣$ . 2. Cross product: $O P \times O Q = 5 \cdot 3 - 7 \cdot 2 7 \cdot 1 - (- 1) \cdot 3 (- 1) \cdot 2 - 5 \cdot 1 = 110 - 7 .$ 3. Magnitude: $1 + 100 + 49 = 150 = 56$ .

Answer: a) $O P = - 1 57$ , $O Q = 123$ ; b) area $= 56$

Exercise 26 BasicIB · P1 TZ0 November 2005

Parallelogram $ABCD$ has vertices $A (3, 2, 0)$ , $B (7, - 1, - 1)$ , $C (10, - 3, 0)$ and $D (6, 0, 1)$ .

Calculate the area of the parallelogram.

Show answer

Answer: area $= 53$

Show working and answer

Two adjacent sides from $A$ : $A B = 4 - 3 - 1$ and $A D = 3 - 2 1$ .
Cross product: $A B \times A D = (- 3) \cdot 1 - (- 1) (- 2) (- 1) \cdot 3 - 4 \cdot 1 4 \cdot (- 2) - (- 3) \cdot 3 = - 5 - 7 1 .$
The area is the magnitude: $25 + 49 + 1 = 75 = 53$ .

Answer: area $= 53$

Exercise 27 MediumIB · P1 TZ2 May 2008Proof

Given any two vectors $a$ and $b$ , both different from $0$ , verify that:

$∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$

View solution

Let $θ$ be the angle between $a$ and $b$ . We know that $∣ a \times b ∣ = ∣ a ∣ ∣ b ∣ sin θ$ and that $a \cdot b = ∣ a ∣ ∣ b ∣ cos θ$ .
We square the vector product and use $sin^{2} θ = 1 - cos^{2} θ$ : $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ = ∣ a ∣^{2} ∣ b ∣^{2} (1 - cos^{2} θ) = ∣ a ∣^{2} ∣ b ∣^{2} - ∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ .$
The last term is $(∣ a ∣ ∣ b ∣ cos θ)^{2} = (a \cdot b)^{2}$ . Therefore $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} .$
(Alternative route: with $a = (u_{1}, u_{2}, u_{3})$ and $b = (v_{1}, v_{2}, v_{3})$ , expanding both sides in components gives the same expression, although it is more laborious.)

Hence proved.

Exercise 28 MediumIB · P1 TZ2 May 2004

Given that $a = (i + 2 j + k) \times (- 2 i + 3 k)$

a) find $a$ .

b) find the vector projection of $a$ onto the vector $- 2 j + k$ .

Show answer

Answer: a) $a = 6 i - 5 j + 4 k$ ; b) $- \frac{28}{5} j + \frac{14}{5} k$

Show working and answer

a) 1. We calculate the vector product: $a = 121 \times - 2 03 = 2 \cdot 3 - 1 \cdot 0 1 \cdot (- 2) - 1 \cdot 3 1 \cdot 0 - 2 \cdot (- 2) = 6 - 5 4 .$ That is, $a = 6 i - 5 j + 4 k$ .

b) 1. Let $d = 0 - 2 1$ . The vector projection of $a$ onto $d$ is $\frac{a \cdot d}{∣ d ∣ ^{2}} d$ . 2. $a \cdot d = 0 + 10 + 4 = 14$ and $∣ d ∣^{2} = 0 + 4 + 1 = 5$ . 3. Therefore the projection is $\frac{14}{5} 0 - 2 1 = 0 - \frac{28}{5} \frac{14}{5}$ .

Answer: a) $a = 6 i - 5 j + 4 k$ ; b) $- \frac{28}{5} j + \frac{14}{5} k$

Exercise 29 MediumIB · P1 TZ1 May 2017

$ABCD$ is a parallelogram, where $A B = - i + 2 j + 3 k$ and $A D = 4 i - j - 2 k$ .

a) Calculate the area of the parallelogram $ABCD$ .

b) Using a suitable scalar product of two vectors, determine whether $ABC$ is an acute or an obtuse angle.

Show answer

Answer: a) area $= 56$ ; b) $ABC$ is acute

Show working and answer

a) 1. The area is $∣ A B \times A D ∣$ . 2. Cross product: $A B \times A D = 2 \cdot (- 2) - 3 \cdot (- 1) 3 \cdot 4 - (- 1) (- 2) (- 1) (- 1) - 2 \cdot 4 = - 1 10 - 7 .$ 3. Magnitude: $1 + 100 + 49 = 150 = 56$ .

b) 1. The angle $A B C$ is the angle between $B A$ and $B C$ . In a parallelogram $B C = A D$ , and $B A = - A B$ . 2. So $B A = 1 - 2 - 3$ and $B C = 4 - 1 - 2$ . 3. Scalar product: $B A \cdot B C = 4 + 2 + 6 = 12 > 0$ . 4. Since the scalar product is positive, $cos A B C > 0$ and the angle is acute.

Answer: a) area $= 56$ ; b) $ABC$ is acute

Exercise 30 MediumIB · P2 TZ2 May 2017Proof

Given that $a \times b = b \times c \neq = 0$ , show that $a + c = s b$ , where $s$ is a scalar.

View solution

We move everything to one side: $a \times b - b \times c = 0$ .
Since $b \times c = - c \times b$ , this becomes $a \times b + c \times b = 0$ .
The cross product is distributive, so $(a + c) \times b = 0$ .
A cross product is zero when one of the two vectors is zero or when they are parallel. Here $b \neq = 0$ , because $a \times b \neq = 0$ . If $a + c = 0$ , we simply take $s = 0$ .
Otherwise, $a + c$ and $b$ are non-zero and parallel, that is, there exists a scalar $s$ with $a + c = s b$ .

As required.

Exercise 31 MediumIB · P1 November 1998

Let the vectors $v = i + 2 j - 3 k$ , $u = 2 i + 2 j + 3 k$ and $w = i + (2 - t) j + (t + 1) k$ be given.

Calculate the value of the parameter $t$ such that the three vectors $u$ , $v$ and $w$ are coplanar.

Show answer

Answer: $t = \frac{8}{7}$

Show working and answer

Three vectors are coplanar if their scalar triple product is zero, that is, if the determinant of their components is $0$ .
We calculate the determinant (expanding along the first row): $211 22 2 - t 3 - 3 t + 1 = 2 [2 (t + 1) + 3 (2 - t)] - 2 [(t + 1) + 3] + 3 [(2 - t) - 2] .$
We simplify: $2 (8 - t) - 2 (t + 4) + 3 (- t) = 16 - 2 t - 2 t - 8 - 3 t = 8 - 7 t$ .
Setting this equal to zero: $8 - 7 t = 0$ , so $t = \frac{8}{7}$ .

Answer: $t = \frac{8}{7}$

Exercise 32 MediumIB · P1 TZ0 May 2006

Let $a = 210$ , $b = - 1 p 6$ and $c = 2 - 4 3$ .

a) Find $a \times b$ .

b) Find the value of $p$ , given that $a \times b$ is parallel to $c$ .

Show answer

Answer: a) $a \times b = 6 i - 12 j + (2 p + 1) k$ ; b) $p = 4$

Show working and answer

a) 1. We apply the formula for the cross product: $a \times b = 1 \cdot 6 - 0 \cdot p 0 \cdot (- 1) - 2 \cdot 6 2 \cdot p - 1 \cdot (- 1) = 6 - 12 2 p + 1 .$

b) 1. Two vectors are parallel if one is a multiple of the other: $a \times b = k c$ . 2. Comparing the first component, $6 = 2 k$ , so $k = 3$ . The second component gives $- 12 = - 4 k$ , which also holds with $k = 3$ . 3. The third component must satisfy $2 p + 1 = 3 k = 9$ , hence $p = 4$ .

Answer: a) $a \times b = 6 i - 12 j + (2 p + 1) k$ ; b) $p = 4$

Exercise 33 MediumIB · P1 TZ2 May 2007

Given that $a = 2 i - j - k$ , $b = 2 i + j - 2 k$ and $c = - i + j - k$ are the position vectors of the points $A$ , $B$ and $C$ respectively, calculate the area of triangle $A B C$ .

Show answer

Answer: Area $= \frac{7}{2}$

Show working and answer

The points are $A (2, - 1, - 1)$ , $B (2, 1, - 2)$ and $C (- 1, 1, - 1)$ .
We calculate two sides from $A$ : $A B = 02 - 1, A C = - 3 20 .$
Their cross product is $A B \times A C = 2 \cdot 0 - (- 1) \cdot 2 (- 1) (- 3) - 0 \cdot 0 0 \cdot 2 - 2 \cdot (- 3) = 236 .$
Its magnitude is $4 + 9 + 36 = 7$ .
The area of the triangle is half the area of the parallelogram: $\frac{1}{2} \cdot 7$ .

Answer: Area $= \frac{7}{2}$

Exercise 34 MediumIB · P1 TZ1 May 2009

A triangle has vertices $A (1, - 1, 1)$ , $B (1, 1, 0)$ and $C (- 1, 1, - 1)$ . Show that the area of the triangle is $6$ .

Show answer

Answer: Area $= 6$ . As required.

Show working and answer

We calculate two sides from $A$ : $A B = 02 - 1, A C = - 2 2 - 2 .$
Cross product: $A B \times A C = 2 \cdot (- 2) - (- 1) \cdot 2 (- 1) (- 2) - 0 \cdot (- 2) 0 \cdot 2 - 2 \cdot (- 2) = - 2 24 .$
Its magnitude is $4 + 4 + 16 = 24 = 26$ .
The area of the triangle is $\frac{1}{2} A B \times A C = \frac{1}{2} \cdot 26 = 6$ .

Answer: Area $= 6$ . As required.

Exercise 35 MediumIB · P2 TZ2 May 2015

Consider the vectors $u = i + 2 j - 2 k$ and $v = a i + b j$ , where $a$ and $b$ are constants.

It is known that $u \times v = 4 i + b j + c k$ , where $c$ is a constant.

a) Find the value of each of the constants $a$ , $b$ and $c$ .

b) Hence, find the Cartesian equation of the plane which contains the vectors $u$ and $v$ and passes through the point $(0, 0, 0)$ .

Show answer

Answer: a) $a = - 1$ , $b = 2$ , $c = 4$ ; b) $2 x + y + 2 z = 0$

Show working and answer

a) 1. We calculate the vector product with $v = (a, b, 0)$ : $u \times v = 2 \cdot 0 - (- 2) \cdot b (- 2) \cdot a - 1 \cdot 0 1 \cdot b - 2 \cdot a = 2 b - 2 a b - 2 a .$ 2. We equate it to $(4, b, c)$ : $2 b = 4$ , so $b = 2$ . 3. $- 2 a = b = 2$ , so $a = - 1$ . 4. $c = b - 2 a = 2 + 2 = 4$ .

b) 1. The vector product $u \times v = (4, 2, 4)$ is normal to the plane, and so is $(2, 1, 2)$ . 2. The plane passes through the origin, so $d = 0$ : $2 x + y + 2 z = 0$ .

Answer: a) $a = - 1$ , $b = 2$ , $c = 4$ ; b) $2 x + y + 2 z = 0$

Exercise 36 MediumIB · P1 TZ2 May 2021Proof

Given two non-zero vectors, $a$ and $b$ , show that $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$

View solution

Let $θ$ be the angle between $a$ and $b$ . The modulus of the cross product is $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$ . Squaring: $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ .$
We use the identity $sin^{2} θ = 1 - cos^{2} θ$ : $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - ∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ .$
By the definition of the scalar product, $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ , so $(a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ$ .
Substituting into step 2: $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} .$

Hence proved.

Exercise 37 HardIB · P1 TZ0 November 2011Proof

a) Verify that, for two non-zero vectors $a$ and $b$ ,

(i) if $∣ a - b ∣ = ∣ a + b ∣$ , then $a$ and $b$ are perpendicular.

(ii) $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$

b) The points $A$ , $B$ and $C$ have position vectors $a$ , $b$ and $c$ .

(i) Verify that the area of triangle $A B C$ is $\frac{1}{2} ∣ a \times b + b \times c + c \times a ∣$

(ii) Hence, verify that the shortest distance between $B$ and $A C$ is $\frac{∣ a \times b + b \times c + c \times a ∣}{∣ c - a ∣}$

View solution

a) i) 1. We square the equality and use $∣ u ∣^{2} = u \cdot u$ : $(a - b) \cdot (a - b) = (a + b) \cdot (a + b) .$ 2. Expanding: $∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2} = ∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2}$ . 3. Simplifying gives $- 2 a \cdot b = 2 a \cdot b$ , that is, $a \cdot b = 0$ . Since both are non-zero, they are perpendicular.

(ii) 1. If $θ$ is the angle between $a$ and $b$ , we know that $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$ and $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ . 2. Then $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ = ∣ a ∣^{2} ∣ b ∣^{2} (1 - cos^{2} θ) = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} .$

b) i) 1. The area of the triangle is $\frac{1}{2} A B \times A C$ , with $A B = b - a$ and $A C = c - a$ . 2. We expand using the fact that the cross product is distributive: $(b - a) \times (c - a) = b \times c - b \times a - a \times c + a \times a .$ 3. Since $a \times a = 0$ , $- b \times a = a \times b$ and $- a \times c = c \times a$ , it follows that $A B \times A C = a \times b + b \times c + c \times a .$ 4. Therefore the area is $\frac{1}{2} ∣ a \times b + b \times c + c \times a ∣$ .

(ii) 1. We take $[A C]$ as the base of the triangle. Its length is $∣ A C ∣ = ∣ c - a ∣$ , and the height is the distance $d$ from $B$ to the line $A C$ . 2. The area is also $\frac{1}{2} ∣ c - a ∣ d$ . 3. Equating with the result of b) i): $\frac{1}{2} ∣ c - a ∣ d = \frac{1}{2} ∣ a \times b + b \times c + c \times a ∣$ . 4. Solving for $d$ gives the required expression.

As required.

Exercise 38 HardIB · P2 November 1999

Consider the points $A (1, 2, 1)$ , $B (0, - 1, 2)$ , $C (1, 0, 2)$ and $D (2, - 1, - 6)$ .

a) Calculate the vectors $A B$ and $B C$ .

b) Calculate $A B \times B C$ .

c) Hence, or otherwise, calculate the area of triangle $A B C$ .

d) Find the equation of the plane $P$ which contains the points $A$ , $B$ and $C$ .

e) Find a set of parametric equations for the line which passes through the point $D$ and is perpendicular to the plane $P$ .

f) Calculate the distance from the point $D$ to the plane $P$ .

g) Calculate a unit vector perpendicular to the plane $P$ .

h) The point $E$ is a reflection of the point $D$ in the plane $P$ . Calculate the coordinates of $E$ .

Show answer

Answer: a) $A B = - 1 - 3 1$ , $B C = 110$ ; b) $- i + j + 2 k$ ; c) $\frac{6}{2}$ ; d) $- x + y + 2 z - 3 = 0$ ; e) $x = 2 - t, y = - 1 + t, z = - 6 + 2 t$ ; f) $36$ ; g) $\frac{1}{6} - 1 12$ ; h) $E (- 4, 5, 6)$

Show working and answer

a) $A B = B - A = - 1 - 3 1$ and $B C = C - B = 110$ .

b) $A B \times B C = (- 3) \cdot 0 - 1 \cdot 1 1 \cdot 1 - (- 1) \cdot 0 (- 1) \cdot 1 - (- 3) \cdot 1 = - 1 12 = - i + j + 2 k .$

c) 1. The area of the triangle is half the modulus of the vector product: $Area = \frac{1}{2} 1 + 1 + 4 = \frac{6}{2}$ .

d) 1. The vector $(- 1, 1, 2)$ is normal to the plane, so the plane is $- x + y + 2 z = d$ . 2. With $A (1, 2, 1)$ : $d = - 1 + 2 + 2 = 3$ . We check with $B$ : $0 - 1 + 4 = 3$ , and with $C$ : $- 1 + 0 + 4 = 3$ . 3. The plane $P$ is $- x + y + 2 z = 3$ , or equivalently $x - y - 2 z + 3 = 0$ .

e) 1. The line has as direction vector the normal $(- 1, 1, 2)$ and passes through $D (2, - 1, - 6)$ : $x = 2 - t, y = - 1 + t, z = - 6 + 2 t .$

f) 1. Using the formula for the distance from a point to a plane, with $- x + y + 2 z - 3 = 0$ : $d (D, P) = \frac{∣ - 2 + ( - 1 ) + 2 ( - 6 ) - 3∣}{1 + 1 + 4} = \frac{18}{6} = 36 .$

g) 1. We divide the normal by its modulus $6$ : $\frac{1}{6} - 1 12$ .

h) 1. We find the foot of the perpendicular, that is, the point of the line in (e) which lies in $P$ : $- (2 - t) + (- 1 + t) + 2 (- 6 + 2 t) = 3$ , that is, $6 t - 15 = 3$ , so $t = 3$ . The foot is $(- 1, 2, 0)$ . 2. The foot is the midpoint of $D$ and $E$ , so $E = 2 (- 1, 2, 0) - (2, - 1, - 6) = (- 4, 5, 6)$ .

Exercise 39 HardIB · P1 TZ2 May 2010

a) Consider the vectors $a = 6 i + 3 j + 2 k$ , $b = - 3 j + 4 k$ .

(i) Find the cosine of the angle between the vectors $a$ and $b$ .

(ii) Find $a \times b$ .

(iii) Hence, find the Cartesian equation of the plane $π$ which contains the vectors $a$ and $b$ and passes through the point $(1, 1, - 1)$ .

(iv) The plane $π$ intersects the $x y$ -plane in the line $L$ . Find the area of the finite triangular region bounded by the line $L$ , the $x$ -axis and the $y$ -axis.

b) Given vectors $p$ and $q$ ,

(i) Verify that $p \cdot p = ∣ p ∣^{2}$ .

(ii) Hence, or otherwise, verify that $∣ p + q ∣^{2} = ∣ p ∣^{2} + 2 p \cdot q + ∣ q ∣^{2}$ .

(iii) Deduce that $∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣$ .

Show answer

Answer: a) i) $- \frac{1}{35}$ ; ii) $18 i - 24 j - 18 k$ ; iii) $3 x - 4 y - 3 z - 2 = 0$ ; iv) $\frac{1}{6}$ . Part b) is proved.

Show working and answer

a) i) 1. $a \cdot b = 6 \cdot 0 + 3 \cdot (- 3) + 2 \cdot 4 = - 1$ . 2. $∣ a ∣ = 36 + 9 + 4 = 7$ and $∣ b ∣ = 9 + 16 = 5$ . 3. $cos θ = \frac{- 1}{7 \cdot 5} = - \frac{1}{35}$ .

a) ii) $a \times b = 3 \cdot 4 - 2 \cdot (- 3) 2 \cdot 0 - 6 \cdot 4 6 \cdot (- 3) - 3 \cdot 0 = 18 - 24 - 18 = 18 i - 24 j - 18 k .$

a) iii) 1. The vector $a \times b$ is normal to the plane, and so is $(3, - 4, - 3)$ (divided by 6). 2. The plane is $3 x - 4 y - 3 z = d$ . With the point $(1, 1, - 1)$ : $d = 3 - 4 + 3 = 2$ .

a) iv) 1. The line $L$ is the intersection with $z = 0$ : $3 x - 4 y = 2$ . 2. It cuts the $x$ -axis ( $y = 0$ ) at $x = \frac{2}{3}$ and the $y$ -axis ( $x = 0$ ) at $y = - \frac{1}{2}$ . 3. The triangle has legs $\frac{2}{3}$ and $\frac{1}{2}$ , so the area is $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}$ .

b) i) 1. By definition, $p \cdot p = ∣ p ∣∣ p ∣ cos 0^{\circ} = ∣ p ∣^{2}$ .

b) ii) 1. We use the distributive property and the commutativity of the dot product: $∣ p + q ∣^{2} = (p + q) \cdot (p + q) = p \cdot p + 2 p \cdot q + q \cdot q = ∣ p ∣^{2} + 2 p \cdot q + ∣ q ∣^{2} .$

b) iii) 1. Since $p \cdot q = ∣ p ∣∣ q ∣ cos θ \leq ∣ p ∣∣ q ∣$ (because $cos θ \leq 1$ ), from b) ii) it follows that $∣ p + q ∣^{2} \leq ∣ p ∣^{2} + 2∣ p ∣∣ q ∣ + ∣ q ∣^{2} = (∣ p ∣ + ∣ q ∣)^{2} .$ 2. Both sides are non-negative, so we can take square roots: $∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣$ .

Answer: a) i) $- \frac{1}{35}$ ; ii) $18 i - 24 j - 18 k$ ; iii) $3 x - 4 y - 3 z - 2 = 0$ ; iv) $\frac{1}{6}$ . Part b) is proved.

Exercise 40 HardIB · P2 TZ2 May 2011

The points $P (- 1, 2, - 3)$ , $Q (- 2, 1, 0)$ , $R (0, 5, 1)$ and $S$ form a parallelogram, with $S$ diagonally opposite to $Q$ .

a) Find the coordinates of $S$ .

b) The vector product $P Q \times P S = - 13 7 m$ . Find the value of $m$ .

c) Hence, calculate the area of the parallelogram $P QR S$ .

d) Find the Cartesian equation of the plane $π_{1}$ which contains the parallelogram $P QR S$ .

e) Write down a vector equation of the line which passes through the origin $(0, 0, 0)$ and is perpendicular to the plane $π_{1}$ .

f) Hence, find the point on the plane which is closest to the origin.

g) A second plane, $π_{2}$ , has equation $x - 2 y + z = 3$ . Calculate the angle between the two planes.

Show answer

Answer: a) $S (1, 6, - 2)$ ; b) $m = - 2$ ; c) $222$ ; d) $13 x - 7 y + 2 z + 33 = 0$ ; e) $r = t 13 - 7 2$ ; f) $(- \frac{143}{74}, \frac{77}{74}, - \frac{11}{37})$ ; g) $37. 4^{\circ}$

Show working and answer

a) 1. In a parallelogram $P QR S$ the diagonals are $P R$ and $QS$ and they bisect each other at their common midpoint, so $P S = QR$ and $S = P + R - Q$ . 2. $S = (- 1 + 0 + 2, 2 + 5 - 1, - 3 + 1 - 0) = (1, 6, - 2)$ .

b) 1. $P Q = (- 1, - 1, 3)$ and $P S = (2, 4, 1)$ . 2. The third component of the vector product is $(- 1) \cdot 4 - (- 1) \cdot 2 = - 2$ . The rest checks: $(- 1) \cdot 1 - 3 \cdot 4 = - 13$ and $3 \cdot 2 - (- 1) \cdot 1 = 7$ . Therefore $m = - 2$ .

c) 1. The area of the parallelogram is the modulus of the vector product: $(- 13)^{2} + 7^{2} + (- 2)^{2} = 169 + 49 + 4 = 222$ .

d) 1. The vector product $(- 13, 7, - 2)$ is normal to the plane: $13 x - 7 y + 2 z = d$ . 2. With $P (- 1, 2, - 3)$ : $d = - 13 - 14 - 6 = - 33$ .

e) 1. The line has as direction vector the normal of the plane: $r = 000 + t 13 - 7 2 .$

f) 1. The point of the plane closest to the origin is the foot of the perpendicular from e): the point of the line that lies in $π_{1}$ . 2. We substitute $(13 t, - 7 t, 2 t)$ into $13 x - 7 y + 2 z = - 33$ : $169 t + 49 t + 4 t = - 33$ , so $222 t = - 33$ and $t = - \frac{11}{74}$ . 3. The point is $(- \frac{143}{74}, \frac{77}{74}, - \frac{22}{74})$ , that is, $(- \frac{143}{74}, \frac{77}{74}, - \frac{11}{37})$ .

g) 1. The angle between two planes is the acute angle between their normals, $n_{1} = (13, - 7, 2)$ and $n_{2} = (1, - 2, 1)$ . 2. $cos θ = \frac{∣13 + 14 + 2∣}{222 6} = \frac{29}{1332}$ . 3. $θ = arccos \frac{29}{1332} \approx 37. 4^{\circ}$ .

Answer: a) $S (1, 6, - 2)$ ; b) $m = - 2$ ; c) $222$ ; d) $13 x - 7 y + 2 z + 33 = 0$ ; e) $r = t 13 - 7 2$ ; f) $(- \frac{143}{74}, \frac{77}{74}, - \frac{11}{37})$ ; g) $37. 4^{\circ}$

Exercise 41 HardIB · P1 TZ2 May 2013

The vertices of a triangle $ABC$ have the following coordinates: $A (- 1, 2, 3)$ , $B (4, 1, 1)$ and $C (3, - 2, 2)$ .

a) (i) Find the length of the sides of the triangle.

(ii) Find $cos BAC$ .

b) (i) Verify that $BC \times CA = - 7 i - 3 j - 16 k$ .

(ii) Hence, verify that the area of the triangle $ABC$ is $\frac{1}{2} 314$ .

c) Find the Cartesian equation of the plane that contains the triangle $ABC$ .

d) Find a vector equation of $(AB)$ .

The point $D$ lies on $(AB)$ , such that $OD$ is perpendicular to $BC$ , where $O$ is the origin.

e) (i) Find the coordinates of $D$ .

(ii) Verify that $D$ does not lie between $A$ and $B$ .

Show answer

Answer: a) (i) $∣ A B ∣ = 30$ , $∣ A C ∣ = 33$ , $∣ B C ∣ = 11$ ; (ii) $cos B A C = \frac{26}{3 110}$ ; c) $7 x + 3 y + 16 z - 47 = 0$ ; d) $r = (- 1, 2, 3) + t (5, - 1, - 2)$ ; e) (i) $D (- \frac{7}{2}, \frac{5}{2}, 4)$ ; (ii) $t = - \frac{1}{2} < 0$ . Parts b) and e) (ii) are verified.

Show working and answer

a) (i) 1. $A B = (5, - 1, - 2)$ , $A C = (4, - 4, - 1)$ , $B C = (- 1, - 3, 1)$ . 2. $∣ A B ∣ = 25 + 1 + 4 = 30$ , $∣ A C ∣ = 16 + 16 + 1 = 33$ , $∣ B C ∣ = 1 + 9 + 1 = 11$ .

(ii) 1. $cos B A C = \frac{A B \cdot A C}{∣ A B ∣∣ A C ∣}$ with $A B \cdot A C = 20 + 4 + 2 = 26$ . 2. $cos B A C = \frac{26}{30 33} = \frac{26}{990} = \frac{26}{3 110}$ .

b) (i) With $C A = A - C = (- 4, 4, 1)$ : $B C \times C A = - 1 - 3 1 \times - 4 41 = (- 3) \cdot 1 - 1 \cdot 4 1 \cdot (- 4) - (- 1) \cdot 1 (- 1) \cdot 4 - (- 3) (- 4) = - 7 - 3 - 16 = - 7 i - 3 j - 16 k .$

(ii) The area of the triangle is half the magnitude of the vector (cross) product: $\frac{1}{2} 49 + 9 + 256 = \frac{1}{2} 314$ .

c) 1. The vector $B C \times C A = (- 7, - 3, - 16)$ is normal to the plane; we use $n = (7, 3, 16)$ . 2. Plane: $7 x + 3 y + 16 z = d$ . With $A (- 1, 2, 3)$ : $d = - 7 + 6 + 48 = 47$ . 3. $7 x + 3 y + 16 z - 47 = 0$ (equivalent to $- 7 x - 3 y - 16 z + 47 = 0$ ).

d) The line passes through $A$ with direction $A B = (5, - 1, - 2)$ : $r = - 1 23 + t 5 - 1 - 2 .$

e) (i) 1. $D = (- 1 + 5 t, 2 - t, 3 - 2 t)$ and $B C = (- 1, - 3, 1)$ . 2. $O D \cdot B C = 0$ : $- (- 1 + 5 t) - 3 (2 - t) + (3 - 2 t) = 0 \Rightarrow 1 - 5 t - 6 + 3 t + 3 - 2 t = 0 \Rightarrow - 2 - 4 t = 0$ . 3. $t = - \frac{1}{2}$ and $D = (- \frac{7}{2}, \frac{5}{2}, 4)$ .

(ii) 1. $A$ corresponds to $t = 0$ and $B$ to $t = 1$ ; the points between $A$ and $B$ are those with $0 \leq t \leq 1$ . 2. $D$ has $t = - \frac{1}{2} < 0$ , so it is not between $A$ and $B$ . Moreover, $D A = \frac{1}{2} A B$ and $D B = \frac{3}{2} A B$ have the same direction.

Exercise 42 HardIB · P1 TZ2 May 2018

Let $a$ , $b$ , $c$ and $d$ be the position vectors, with respect to the origin O, of the points A, B, C and D respectively. It is known that $A B = D C$ .

a) i) Explain why ABCD is a parallelogram.

(ii) Using vector algebra, show that $A D = B C$ .

The position vectors $O A$ , $O B$ , $O C$ and $O D$ are given by

$a = i + 2 j - 3 k$

$b = 3 i - j + p k$

$c = q i + j + 2 k$

$d = - i + r j - 2 k$

where $p$ , $q$ and $r$ are constants.

b) Show that $p = 1$ , $q = 1$ and $r = 4$ .

c) Find the area of the parallelogram ABCD.

The point at which the diagonals of ABCD intersect is called M.

d) Find the vector equation of the line that passes through M and is normal to the plane $π$ that contains ABCD.

e) Find the Cartesian equation of $π$ .

The plane $π$ cuts the $x$ , $y$ and $z$ axes at X, Y and Z respectively.

f) i) Find the coordinates of X, Y and Z.

(ii) Find YZ.

Show answer

Answer: b) $p = 1$ , $q = 1$ , $r = 4$ ; c) $15$ ; d) $r = 1 \frac{3}{2} - \frac{1}{2} + t - 11 - 10 - 2$ ; e) $11 x + 10 y + 2 z = 25$ ; f) i) $X (\frac{25}{11}, 0, 0)$ , $Y (0, \frac{5}{2}, 0)$ , $Z (0, 0, \frac{25}{2})$ ; ii) $Y Z = \frac{5 26}{2}$ . Part a) consists of explanation and proof.

Show working and answer

a) i) 1. $A B = D C$ means that the segments $[A B]$ and $[D C]$ have the same magnitude, the same direction and the same sense. Two opposite sides that are parallel and equal are enough for the quadrilateral to be a parallelogram.

(ii) 1. We go from A to D via B and C: $A D = A B + B C + C D$ . 2. As $C D = - D C = - A B$ , we get $A D = A B + B C - A B = B C$ .

b) 1. $A B = b - a = 2 - 3 p + 3$ and $D C = c - d = q + 1 1 - r 4$ . 2. Equating component by component: $q + 1 = 2$ , $1 - r = - 3$ and $p + 3 = 4$ . 3. Hence $q = 1$ , $r = 4$ and $p = 1$ .

c) 1. The area of the parallelogram is $∣ A B \times A D ∣$ , with $A B = 2 - 3 4$ and $A D = d - a = - 2 21$ . 2. $A B \times A D = (- 3) (1) - (4) (2) (4) (- 2) - (2) (1) (2) (2) - (- 3) (- 2) = - 11 - 10 - 2 .$ 3. Area $= 121 + 100 + 4 = 225 = 15$ .

d) 1. The diagonals of a parallelogram bisect each other, so M is the midpoint of $[A C]$ : $M = \frac{A + C}{2} = \frac{( 1 , 2 , - 3 ) + ( 1 , 1 , 2 )}{2} = (1, \frac{3}{2}, - \frac{1}{2})$ . 2. A normal vector to $π$ is $A B \times A D = (- 11, - 10, - 2)$ , which is the direction vector of the required line. 3. $r = 1 \frac{3}{2} - \frac{1}{2} + t - 11 - 10 - 2, t \in R .$

e) 1. With normal $(11, 10, 2)$ (the opposite of the previous one), the plane is $11 x + 10 y + 2 z = d$ . 2. Through $A (1, 2, - 3)$ : $d = 11 + 20 - 6 = 25$ . 3. $π : 11 x + 10 y + 2 z = 25$ , that is, $- 11 x - 10 y - 2 z + 25 = 0$ .

f) i) 1. $x$ -axis ( $y = z = 0$ ): $11 x = 25$ , $X (\frac{25}{11}, 0, 0)$ . 2. $y$ -axis ( $x = z = 0$ ): $10 y = 25$ , $Y (0, \frac{5}{2}, 0)$ . 3. $z$ -axis ( $x = y = 0$ ): $2 z = 25$ , $Z (0, 0, \frac{25}{2})$ .

(ii) 1. $Y Z = Z - Y = (0, - \frac{5}{2}, \frac{25}{2})$ . 2. $Y Z = \frac{25}{4} + \frac{625}{4} = \frac{650}{4} = \frac{5 26}{2}$ .

Exercise 43 HardIB · P2 TZ0 May 2006

Let $A$ be the point $(2, - 1, 0)$ , $B$ the point $(3, 0, 1)$ and $C$ the point $(1, m, 2)$ , where $m \in Z$ , $m < 0$ .

a) i) Find the scalar (dot) product $B A \cdot B C$

(ii) Hence, given that $A B C = arccos \frac{2}{3}$ , verify that $m = - 1$ .

b) Determine the Cartesian equation of the plane $A B C$ .

c) Find the area of triangle $A B C$ .

d) i) The line $L$ is perpendicular to the plane $A B C$ and passes through $A$ . Find a vector equation of $L$ .

(ii) The point $D (6, - 7, 2)$ lies on $L$ . Find the volume of the pyramid $A B C D$ .

Show answer

Answer: a) i) $1 - m$ , ii) $m = - 1$ ; b) $2 x - 3 y + z = 7$ (that is, $- 2 x + 3 y - z + 7 = 0$ ); c) $\frac{14}{2}$ ; d) i) $r = 2 - 1 0 + t 2 - 3 1$ , ii) $V = \frac{14}{3}$

Show working and answer

a) i) 1. $B A = A - B = - 1 - 1 - 1$ and $B C = C - B = - 2 m 1$ . 2. $B A \cdot B C = (- 1) (- 2) + (- 1) m + (- 1) (1) = 2 - m - 1 = 1 - m$ .

(ii) 1. The magnitudes are $∣ B A ∣ = 3$ and $∣ B C ∣ = 5 + m^{2}$ . Then $cos A B C = \frac{1 - m}{3 5 + m ^{2}} = \frac{2}{3} .$ 2. Square both sides: $\frac{( 1 - m ) ^{2}}{3 ( 5 + m ^{2} )} = \frac{2}{9} \Rightarrow 9 (1 - 2 m + m^{2}) = 6 (5 + m^{2}) \Rightarrow 3 m^{2} - 18 m - 21 = 0 \Rightarrow m^{2} - 6 m - 7 = 0$ . 3. The solutions are $m = 7$ and $m = - 1$ . Since $m < 0$ , $m = - 1$ . (Also, for $m = - 1$ the scalar product $1 - m = 2$ is positive, as is $cos A B C$ .)

b) 1. With $m = - 1$ : $C (1, - 1, 2)$ . Two vectors in the plane are $A B = (1, 1, 1)$ and $A C = (- 1, 0, 2)$ . 2. Their vector (cross) product gives the normal: $A B \times A C = 1 \cdot 2 - 1 \cdot 0 1 \cdot (- 1) - 1 \cdot 2 1 \cdot 0 - 1 \cdot (- 1) = 2 - 3 1 .$ 3. The plane is $2 x - 3 y + z = d$ ; with $A (2, - 1, 0)$ : $d = 4 + 3 + 0 = 7$ . 4. $A B C : 2 x - 3 y + z = 7$ .

c) 1. The area of the triangle is half the magnitude of the vector (cross) product: $Area = \frac{1}{2} ∣ A B \times A C ∣ = \frac{1}{2} 4 + 9 + 1 = \frac{14}{2} .$

d) i) 1. The line perpendicular to the plane $A B C$ has the normal $(2, - 3, 1)$ as its direction and passes through $A$ : $L : r = 2 - 1 0 + t 2 - 3 1 .$

(ii) 1. $A D = (4, - 6, 2) = 2 (2, - 3, 1)$ , so $D \in L$ (with $t = 2$ ) and $A D$ is perpendicular to the plane $A B C$ : the height of the pyramid is $h = ∣ A D ∣ = 214$ . 2. The volume is one third of the base area times the height: $V = \frac{1}{3} \cdot \frac{14}{2} \cdot 214 = \frac{14}{3} .$

Answer: a) i) $1 - m$ , ii) $m = - 1$ ; b) $2 x - 3 y + z = 7$ (that is, $- 2 x + 3 y - z + 7 = 0$ ); c) $\frac{14}{2}$ ; d) i) $r = 2 - 1 0 + t 2 - 3 1$ , ii) $V = \frac{14}{3}$

Exercise 44 HardIB · P2 TZ0 November 2010

The following diagram shows a cube $O A B C D E F G$ .

Let $O$ be the origin, $(O A)$ the $x$ -axis, $(O C)$ the $y$ -axis and $(O D)$ the $z$ -axis.

Let $M$ , $N$ and $P$ be the midpoints of $[F G]$ , $[D G]$ and $[C G]$ , respectively.

The coordinates of $F$ are $(2, 2, 2)$ .

a) Find the position vectors $O M$ , $O N$ and $O P$ in terms of their components.

b) Find $M P \times M N$ .

c) Hence,

(i) calculate the area of triangle $M N P$ .

(ii) verify that the line $(A G)$ is perpendicular to the plane $M N P$ .

(iii) find the equation of the plane $M N P$ .

d) Determine the coordinates of the point where the line $(A G)$ meets the plane $M N P$ .

Show answer

Answer: a) $O M = 122$ , $O N = 012$ , $O P = 021$ . b) $- i + j + k$ . c) (i) $\frac{3}{2}$ ; (ii) $A G = 2 (M P \times M N)$ ; (iii) $- x + y + z - 3 = 0$ . d) $(\frac{1}{3}, \frac{5}{3}, \frac{5}{3})$ .

Show working and answer

a) 1. The coordinates of the vertices of the cube are $A (2, 0, 0)$ , $C (0, 2, 0)$ , $D (0, 0, 2)$ and $G (0, 2, 2)$ (from $O$ and $F$ we deduce that the edge is $2$ ). 2. $M$ is the midpoint of $F (2, 2, 2)$ and $G (0, 2, 2)$ : $M = (1, 2, 2)$ . 3. $N$ is the midpoint of $D (0, 0, 2)$ and $G (0, 2, 2)$ : $N = (0, 1, 2)$ . 4. $P$ is the midpoint of $C (0, 2, 0)$ and $G (0, 2, 2)$ : $P = (0, 2, 1)$ . $O M = 122, O N = 012, O P = 021 .$

b) 1. $M P = P - M = - 1 0 - 1$ and $M N = N - M = - 1 - 1 0$ . 2. Their cross product is $M P \times M N = 0 \cdot 0 - (- 1) (- 1) (- 1) (- 1) - (- 1) \cdot 0 (- 1) (- 1) - 0 \cdot (- 1) = - 1 11 = - i + j + k .$

c) (i) The area of the triangle is half the modulus of the cross product: $\frac{1}{2} 1 + 1 + 1 = \frac{3}{2}$ .

(ii) 1. $A G = G - A = - 2 22 = 2 - 1 11 = 2 (M P \times M N)$ . 2. The cross product is perpendicular to the plane $M N P$ , and $A G$ is parallel to it. Therefore $(A G)$ is perpendicular to the plane $M N P$ .

(iii) 1. A normal to the plane is $- 1 11$ , so $- x + y + z = d$ . 2. With $M (1, 2, 2)$ : $d = - 1 + 2 + 2 = 3$ . 3. Plane $M N P$ : $- x + y + z = 3$ , that is, $- x + y + z - 3 = 0$ .

d) 1. The line $(A G)$ is $(x, y, z) = (2 - 2 s, 2 s, 2 s)$ . 2. Substituting into the plane: $- (2 - 2 s) + 2 s + 2 s = 3$ , that is, $- 2 + 6 s = 3$ , so $s = \frac{5}{6}$ . 3. The point is $(2 - \frac{5}{3}, \frac{5}{3}, \frac{5}{3}) = (\frac{1}{3}, \frac{5}{3}, \frac{5}{3})$ .

Exercise 45 HardIB · P1 TZ0 November 2013

Consider the points $A (1, 0, 0)$ , $B (2, 2, 2)$ and $C (0, 2, 1)$ .

a) Find the vector $C A \times C B$ .

b) Find an exact value for the area of triangle $A B C$ .

c) Verify that the Cartesian equation of $Π_{1}$ , the plane containing triangle $A B C$ , is $2 x + 3 y - 4 z = 2$ .

A second plane $Π_{2}$ is defined by the Cartesian equation $Π_{2} : 4 x - y - z = 4$ .

$L_{1}$ is the line of intersection of the planes $Π_{1}$ and $Π_{2}$ .

d) Find a vector equation of $L_{1}$ .

A third plane $Π_{3}$ is defined by the Cartesian equation $16 x + α y - 3 z = β$ .

e) Find the value of $α$ for which the three planes contain the line $L_{1}$ .

f) Find the conditions that $α$ and $β$ must satisfy so that the plane $Π_{3}$ does not intersect $L_{1}$ .

Show answer

Answer: a) $C A \times C B = - 2 i - 3 j + 4 k$ ; b) $\frac{29}{2}$ ; d) $r = 0 - 2 - 2 + λ 122$ ; e) $α = - 5$ ; f) $α = - 5$ and $β \neq = 16$ .

Show working and answer

a) 1. We calculate the two vectors with origin at $C$ : $C A = A - C = 1 - 2 - 1$ and $C B = B - C = 201$ . 2. Cross product: $C A \times C B = (- 2) (1) - (- 1) (0) (- 1) (2) - (1) (1) (1) (0) - (- 2) (2) = - 2 - 3 4 = - 2 i - 3 j + 4 k .$

b) 1. The area of the triangle is half the modulus of the cross product: $Area = \frac{1}{2} C A \times C B$ . 2. $C A \times C B = 4 + 9 + 16 = 29$ , so the area is $\frac{29}{2}$ .

c) 1. A normal vector to the plane is $C A \times C B$ , parallel to $(2, 3, - 4)$ . Therefore the plane has the form $2 x + 3 y - 4 z = d$ . 2. Substituting the three points: $A$ : $2 \cdot 1 = 2$ ; $B$ : $4 + 6 - 8 = 2$ ; $C$ : $0 + 6 - 4 = 2$ . All three satisfy $2 x + 3 y - 4 z = 2$ and are not collinear (the cross product is not zero), so this is the equation of the plane $Π_{1}$ .

d) 1. The direction of the line is perpendicular to the two normal vectors, $n_{1} = (2, 3, - 4)$ and $n_{2} = (4, - 1, - 1)$ : $n_{1} \times n_{2} = 3 (- 1) - (- 4) (- 1) (- 4) (4) - 2 (- 1) 2 (- 1) - 3 \cdot 4 = - 7 - 14 - 14 = - 7122 .$ 2. Look for a common point by setting $x = 0$ in the system: $3 y - 4 z = 2$ and $- y - z = 4$ . From the second, $y = - 4 - z$ ; then $- 12 - 3 z - 4 z = 2$ , so $z = - 2$ and $y = - 2$ . The point is $(0, - 2, - 2)$ . 3. A vector equation is $L_{1} : r = 0 - 2 - 2 + λ 122 .$

e) 1. For $Π_{3}$ to contain $L_{1}$ , the direction vector of the line must be perpendicular to the normal of $Π_{3}$ , $n_{3} = (16, α, - 3)$ : $(1, 2, 2) \cdot (16, α, - 3) = 16 + 2 α - 6 = 0 \Rightarrow α = - 5.$ 2. In addition, the point $(0, - 2, - 2)$ of the line must lie on the plane: $- 5 (- 2) - 3 (- 2) = 16 = β$ . With $α = - 5$ and $β = 16$ the three planes contain $L_{1}$ .

f) 1. The line and the plane do not intersect if they are parallel and the line is not contained in the plane. 2. Parallel: $α = - 5$ (the same condition as in e). If $α \neq = - 5$ the line meets the plane at one point. 3. Not contained: the point $(0, - 2, - 2)$ does not satisfy the equation, that is, $β \neq = 16$ .

Answer: a) $C A \times C B = - 2 i - 3 j + 4 k$ ; b) $\frac{29}{2}$ ; d) $r = 0 - 2 - 2 + λ 122$ ; e) $α = - 5$ ; f) $α = - 5$ and $β \neq = 16$ .

Equations of a line

Exercise 46 MediumIB · P1 November 2001

The point $A (3, 0, - 2)$ lies on the line $r = 3 i - 2 k + λ (2 i - 2 j + k)$ , where $λ$ is a real parameter. Calculate the coordinates of a point on the line that is 6 units from $A$ .

Show answer

Answer: $(7, - 4, 0)$ or $(- 1, 4, - 4)$

Show working and answer

$A$ corresponds to $λ = 0$ . A point on the line is $X (3 + 2 λ, - 2 λ, - 2 + λ)$ , and $A X = λ 2 - 2 1$ .
Its magnitude is $∣ λ ∣ 4 + 4 + 1 = 3∣ λ ∣$ .
We impose $3∣ λ ∣ = 6$ , so $λ = 2$ or $λ = - 2$ .
With $λ = 2$ : $(7, - 4, 0)$ . With $λ = - 2$ : $(- 1, 4, - 4)$ .

Answer: $(7, - 4, 0)$ or $(- 1, 4, - 4)$

Exercise 47 MediumIB · P2 November 2002

The points $A (1, 3, - 17)$ and $B (6, - 7, 8)$ lie on the line $L$ .

a) Find an equation of the line $L$ , giving your answer in parametric form.

b) The point $P$ lies on $L$ and is such that $O P$ is perpendicular to $L$ . Find the coordinates of $P$ .

Show answer

Answer: a) $x = 1 + t, y = 3 - 2 t, z = - 17 + 5 t$ ; b) $P (4, - 3, - 2)$

Show working and answer

a) 1. Direction vector: $A B = 5 - 10 25$ , which is proportional to $d = 1 - 2 5$ . 2. Parametric equations from $A$ : $x = 1 + t$ , $y = 3 - 2 t$ , $z = - 17 + 5 t$ .

b) 1. A point on $L$ is $P (1 + t, 3 - 2 t, - 17 + 5 t)$ , and $O P$ is its position vector. 2. $O P \cdot d = 0$ : $(1 + t) - 2 (3 - 2 t) + 5 (- 17 + 5 t) = 0$ . 3. Expanding: $1 + t - 6 + 4 t - 85 + 25 t = 30 t - 90 = 0$ , so $t = 3$ . 4. $P (4, - 3, - 2)$ .

Answer: a) $x = 1 + t, y = 3 - 2 t, z = - 17 + 5 t$ ; b) $P (4, - 3, - 2)$

Exercise 48 MediumIB · P1 TZ2 May 2008

Consider the points $A (1, - 1, 4)$ , $B (2, - 2, 5)$ and $O (0, 0, 0)$ .

a) Calculate the cosine of the angle between $O A$ and $A B$ .

b) Find a vector equation of the line $L_{1}$ which passes through $A$ and $B$ .

The line $L_{2}$ has equation $r = 2 i + 4 j + 7 k + t (2 i + j + 3 k)$ , where $t \in R$ .

c) Verify that the lines $L_{1}$ and $L_{2}$ intersect, and find the coordinates of the point of intersection.

d) Find the Cartesian equation of the plane which contains both the line $L_{2}$ and the point $A$ .

Show answer

Answer: a) $cos θ = \frac{6}{3}$ ; b) $r = 1 - 1 4 + s 1 - 1 1$ ; c) $(- 2, 2, 1)$ ; d) $4 x + y - 3 z + 9 = 0$

Show working and answer

a) 1. $O A = 1 - 1 4$ and $A B = B - A = 1 - 1 1$ . 2. Dot product: $O A \cdot A B = 1 + 1 + 4 = 6$ . Moduli: $∣ O A ∣ = 18 = 32$ and $∣ A B ∣ = 3$ . 3. $cos θ = \frac{6}{3 2 \cdot 3} = \frac{2}{6} = \frac{6}{3}$ .

b) 1. We take $A$ as a point and $A B$ as direction vector: $r = 1 - 1 4 + s 1 - 1 1 .$

c) 1. We equate the coordinates of $L_{1}$ and $L_{2}$ : $1 + s = 2 + 2 t, - 1 - s = 4 + t, 4 + s = 7 + 3 t .$ 2. Adding the first two: $0 = 6 + 3 t$ , so $t = - 2$ . From the first, $s = 1 + 2 t = - 3$ . 3. We check with the third equation: $4 + (- 3) = 1$ and $7 + 3 (- 2) = 1$ . It holds, so the lines intersect. 4. The point is obtained with $t = - 2$ in $L_{2}$ : $(2 - 4, 4 - 2, 7 - 6) = (- 2, 2, 1)$ .

d) 1. The plane contains $L_{2}$ , with direction vector $d = (2, 1, 3)$ , and passes through $(2, 4, 7)$ . A second vector in the plane is the one from $(2, 4, 7)$ to $A$ : $w = (- 1, - 5, - 3)$ . 2. A normal vector is $d \times w = 1 \cdot (- 3) - 3 \cdot (- 5) 3 \cdot (- 1) - 2 \cdot (- 3) 2 \cdot (- 5) - 1 \cdot (- 1) = 123 - 9 ∥ 41 - 3 .$ 3. The plane is $4 x + y - 3 z = d$ . With $A (1, - 1, 4)$ : $d = 4 - 1 - 12 = - 9$ .

Answer: a) $cos θ = \frac{6}{3}$ ; b) $r = 1 - 1 4 + s 1 - 1 1$ ; c) $(- 2, 2, 1)$ ; d) $4 x + y - 3 z + 9 = 0$

Exercise 49 MediumIB · P1 TZ0 November 2017

The points $A$ and $B$ are given by $A (0, 3, - 6)$ and $B (6, - 5, 11)$ . The plane $π$ is defined by the equation $4 x - 3 y + 2 z = 20$ .

a) Calculate the vector equation of the line $L$ which passes through the points $A$ and $B$ .

b) Calculate the coordinates of the point of intersection of the line $L$ and the plane $π$ .

Show answer

Answer: a) $r = 03 - 6 + t 6 - 8 17$ ; b) $(3, - 1, \frac{5}{2})$

Show working and answer

a) 1. A direction vector of $L$ is $A B = B - A = 6 - 8 17$ . 2. Taking $A$ as a point on the line: $L : r = 03 - 6 + t 6 - 8 17, t \in R .$

b) 1. A point of $L$ has coordinates $(6 t, 3 - 8 t, - 6 + 17 t)$ . 2. For it to lie on $π$ , its coordinates must satisfy its equation: $4 (6 t) - 3 (3 - 8 t) + 2 (- 6 + 17 t) = 20.$ 3. Expanding: $24 t - 9 + 24 t - 12 + 34 t = 20$ , that is, $82 t - 21 = 20$ , from which $t = \frac{1}{2}$ . 4. Substituting $t = \frac{1}{2}$ into the line: $x = 3$ , $y = 3 - 4 = - 1$ , $z = - 6 + \frac{17}{2} = \frac{5}{2}$ .

Answer: a) $r = 03 - 6 + t 6 - 8 17$ ; b) $(3, - 1, \frac{5}{2})$

Exercise 50 HardIB · P2 November 2001

a) Solve the following system of linear equations $x + 3 y - 2 z 2 x + y + 3 z 3 x - y + z = - 6 = 7 = 6$

b) Calculate the vector $v = (i + 3 j - 2 k) \times (2 i + j + 3 k)$ .

c) If $a = i + 3 j - 2 k$ , $b = 2 i + j + 3 k$ and $u = m a + n b$ , where $m$ and $n$ are scalars and $u \neq = 0$ , show that $v$ is perpendicular to $u$ for all $m$ and $n$ .

d) The line $L$ lies in the plane $3 x - y + z = 6$ , passes through the point $(1, - 1, 2)$ and is perpendicular to $v$ . Find the equation of $L$ .

Show answer

Answer: a) $x = 1$ , $y = - 1$ , $z = 2$ ; b) $v = 11 i - 7 j - 5 k$ ; c) $v \cdot u = 0$ ; d) $L : \frac{x - 1}{6} = \frac{y + 1}{13} = \frac{z - 2}{- 5}$

Show working and answer

a) 1. Adding the second and third equations: $5 x + 4 z = 13$ . 2. Multiplying the third equation by 3 and adding it to the first: $10 x + z = 12$ , so $z = 12 - 10 x$ . 3. Substituting into $5 x + 4 z = 13$ : $5 x + 48 - 40 x = 13$ , so $- 35 x = - 35$ and $x = 1$ . Then $z = 2$ . 4. From the third equation: $3 - y + 2 = 6$ , so $y = - 1$ . 5. Check in the first equation: $1 - 3 - 4 = - 6$ . Correct. (On Paper 2 this can also be solved using the GDC.)

b) $v = 13 - 2 \times 213 = 3 \cdot 3 - (- 2) \cdot 1 (- 2) \cdot 2 - 1 \cdot 3 1 \cdot 1 - 3 \cdot 2 = 11 - 7 - 5 .$

c) 1. $v = a \times b$ is perpendicular to both $a$ and $b$ , so $v \cdot a = 0$ and $v \cdot b = 0$ . 2. Then $v \cdot u = m (v \cdot a) + n (v \cdot b) = 0 + 0 = 0$ for any $m$ and $n$ . 3. Therefore $v ⊥ u$ .

d) 1. The direction vector $d$ of $L$ must be perpendicular to $v$ and lie in the plane, that is, perpendicular to the normal of the plane $n = (3, - 1, 1)$ . 2. Therefore $d = v \times n = (- 7) \cdot 1 - (- 5) (- 1) (- 5) \cdot 3 - 11 \cdot 1 11 \cdot (- 1) - (- 7) \cdot 3 = - 12 - 26 10$ , which is proportional to $613 - 5$ . 3. $L$ passes through $(1, - 1, 2)$ , which satisfies $3 + 1 + 2 = 6$ (so it lies in the plane). 4. $L : \frac{x - 1}{6} = \frac{y + 1}{13} = \frac{z - 2}{- 5}$ .

Answer: a) $x = 1$ , $y = - 1$ , $z = 2$ ; b) $v = 11 i - 7 j - 5 k$ ; c) $v \cdot u = 0$ ; d) $L : \frac{x - 1}{6} = \frac{y + 1}{13} = \frac{z - 2}{- 5}$

Exercise 51 HardIB · P2 November 2003

The line $L$ passes through the point $A (2, 5, - 1)$ and is perpendicular to the plane with equation $x + y + z - 1 = 0$ .

a) Find the Cartesian equation of the line $L$ .

b) Find the point of intersection of the line $L$ and the plane.

c) Find the coordinates of the reflection of $A$ in the plane.

d) Calculate the distance from the point $B (2, 0, 6)$ to the line $L$ .

Show answer

Answer: a) $L : \frac{x - 2}{1} = \frac{y - 5}{1} = \frac{z + 1}{1}$ ; b) $(\frac{1}{3}, \frac{10}{3}, - \frac{8}{3})$ ; c) $(- \frac{4}{3}, \frac{5}{3}, - \frac{13}{3})$ ; d) $\frac{654}{3} \approx 8.52$

Show working and answer

a) 1. The direction vector of $L$ is the normal of the plane, $d = (1, 1, 1)$ . 2. $L : \frac{x - 2}{1} = \frac{y - 5}{1} = \frac{z + 1}{1}$ .

b) 1. Points of $L$ : $(2 + t, 5 + t, - 1 + t)$ . 2. We substitute into the plane: $(2 + t) + (5 + t) + (- 1 + t) - 1 = 0$ , that is, $5 + 3 t = 0$ , so $t = - \frac{5}{3}$ . 3. The point is $M (\frac{1}{3}, \frac{10}{3}, - \frac{8}{3})$ .

c) 1. The reflected point $A^{'}$ satisfies that $M$ is the midpoint of $[A A^{'}]$ , so $A^{'} = 2 M - A$ . 2. $A^{'} = (\frac{2}{3} - 2, \frac{20}{3} - 5, - \frac{16}{3} + 1) = (- \frac{4}{3}, \frac{5}{3}, - \frac{13}{3})$ .

d) 1. $B A = 05 - 7$ and $d = (1, 1, 1)$ with $∣ d ∣ = 3$ . 2. $B A \times d = 5 \cdot 1 - (- 7) \cdot 1 (- 7) \cdot 1 - 0 \cdot 1 0 \cdot 1 - 5 \cdot 1 = 12 - 7 - 5, B A \times d = 144 + 49 + 25 = 218 .$ 3. $d (B, L) = \frac{218}{3} = \frac{218}{3} = \frac{654}{3} \approx 8.52$ .

Exercise 52 HardIB · P2 TZ1 May 2013

Consider the points $P (- 3, - 1, 2)$ and $Q (5, 5, 6)$ .

a) Find a vector equation of the line $L_{1}$ which passes through the points $P$ and $Q$ .

The line $L_{2}$ has equation $r = - 4 04 + s 520$ .

b) Show that $L_{1}$ and $L_{2}$ intersect at the point $R (1, 2, 4)$ .

c) Find the acute angle between $L_{1}$ and $L_{2}$ .

Let $S$ be a point on $L_{2}$ such that $∣ RP ∣ = ∣ RS ∣$ .

d) Show that one possible position for $S$ is $S_{1} (- 4, 0, 4)$ and find the coordinates of the other possible position $S_{2}$ .

e) Find a vector equation of the line which passes through $R$ and bisects $PR S_{1}$ .

Show answer

Answer: a) $r = (- 3, - 1, 2) + t (8, 6, 4)$ ; c) $26. 3^{\circ}$ ; d) $S_{2} (6, 4, 4)$ ; e) $r = (1, 2, 4) + λ (- 9, - 5, - 2)$ . Parts b) and the first part of d) are proved.

Show working and answer

a) 1. Direction vector: $P Q = Q - P = 864$ . 2. $L_{1} : r = - 3 - 1 2 + t 864$ .

b) 1. On $L_{1}$ : with $t = \frac{1}{2}$ we obtain $(- 3 + 4, - 1 + 3, 2 + 2) = (1, 2, 4)$ . 2. On $L_{2}$ : with $s = 1$ we obtain $(- 4 + 5, 0 + 2, 4 + 0) = (1, 2, 4)$ . 3. $R$ belongs to both lines and, as they are not parallel, they intersect only at $R$ .

c) 1. $d_{1} = (8, 6, 4)$ , $d_{2} = (5, 2, 0)$ ; $d_{1} \cdot d_{2} = 40 + 12 = 52$ . 2. $∣ d_{1} ∣ = 116 = 229$ , $∣ d_{2} ∣ = 29$ . 3. $cos θ = \frac{52}{2 29 29} = \frac{26}{29}$ , so $θ \approx 26. 3^{\circ}$ .

d) 1. $R P = P - R = (- 4, - 3, - 2)$ , of magnitude $29$ . 2. A point on $L_{2}$ is $S = (- 4 + 5 s, 2 s, 4)$ , and $R S = (5 s - 5, 2 s - 2, 0) = (s - 1) (5, 2, 0)$ . 3. $∣ R S ∣ = ∣ s - 1∣ 29$ . Equating it to $29$ : $∣ s - 1∣ = 1$ , that is $s = 0$ or $s = 2$ . 4. With $s = 0$ we obtain $S_{1} (- 4, 0, 4)$ , as required. With $s = 2$ we obtain $S_{2} (6, 4, 4)$ .

e) 1. Triangle $P R S_{1}$ is isosceles with $∣ R P ∣ = ∣ R S_{1} ∣ = 29$ ; the bisector of the angle at $R$ has the direction of $R P + R S_{1}$ (the diagonal of the rhombus they form). 2. $R S_{1} = (- 5, - 2, 0)$ , so $R P + R S_{1} = (- 9, - 5, - 2)$ . 3. $r = 124 + λ - 9 - 5 - 2$ .

Answer: a) $r = (- 3, - 1, 2) + t (8, 6, 4)$ ; c) $26. 3^{\circ}$ ; d) $S_{2} (6, 4, 4)$ ; e) $r = (1, 2, 4) + λ (- 9, - 5, - 2)$ . Parts b) and the first part of d) are proved.

Exercise 53 HardIB · P1 TZ2 May 2014

Given the points $A (1, 0, 4)$ , $B (2, 3, - 1)$ and $C (0, 1, - 2)$ ,

a) Find the vector equation of the line $L_{1}$ which passes through the points $A$ and $B$ .

The line $L_{2}$ has Cartesian equation $\frac{x - 1}{3} = \frac{y + 2}{1} = \frac{z - 1}{- 2}$ .

b) Show that $L_{1}$ and $L_{2}$ are skew lines.

Consider the plane $Π_{1}$ , parallel to $L_{1}$ and also to $L_{2}$ . The point $C$ lies on the plane $Π_{1}$ .

c) Find the Cartesian equation of the plane $Π_{1}$ .

The line $L_{3}$ has vector equation $r = 301 + λ k 1 - 1$ .

The plane $Π_{2}$ has Cartesian equation $x + y = 12$ .

The angle between the line $L_{3}$ and the plane $Π_{2}$ is equal to $6 0^{\circ}$ .

d) (i) Find the value of $k$ .

(ii) Find the point of intersection $P$ of the line $L_{3}$ and the plane $Π_{2}$ .

Show answer

Answer: a) $r = (1, 0, 4) + t (1, 3, - 5)$ ; c) $x + 13 y + 8 z + 3 = 0$ ; d) (i) $k = 2$ ; (ii) $P (9, 3, - 2)$ . Part b) is proved.

Show working and answer

a) 1. $A B = (1, 3, - 5)$ . 2. $L_{1} : r = 104 + t 13 - 5$ .

b) 1. Directions $d_{1} = (1, 3, - 5)$ and $d_{2} = (3, 1, - 2)$ : they are not proportional, so the lines are not parallel. 2. $L_{2}$ passes through $(1, - 2, 1)$ : $(x, y, z) = (1 + 3 s, - 2 + s, 1 - 2 s)$ . Equating it with $L_{1}$ : $1 + t = 1 + 3 s, 3 t = - 2 + s, 4 - 5 t = 1 - 2 s .$ 3. From the first, $t = 3 s$ ; in the second, $9 s = - 2 + s$ , so $s = - \frac{1}{4}$ and $t = - \frac{3}{4}$ . 4. In the third: $4 + \frac{15}{4} = \frac{31}{4}$ versus $1 + \frac{1}{2} = \frac{3}{2}$ . They do not coincide, so there is no common point. 5. Neither parallel nor intersecting: the lines are skew.

c) 1. Normal: $d_{1} \times d_{2} = 3 \cdot (- 2) - (- 5) \cdot 1 (- 5) \cdot 3 - 1 \cdot (- 2) 1 \cdot 1 - 3 \cdot 3 = - 1 - 13 - 8$ ; we use $(1, 13, 8)$ . 2. $Π_{1} : x + 13 y + 8 z = d$ with $C (0, 1, - 2)$ : $d = 13 - 16 = - 3$ . 3. $Π_{1} : x + 13 y + 8 z + 3 = 0$ .

d) (i) 1. Direction of $L_{3}$ : $d = (k, 1, - 1)$ . Normal of $Π_{2}$ : $n = (1, 1, 0)$ . 2. The angle $θ$ between a line and a plane satisfies $sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣}$ ; with $θ = 6 0^{\circ}$ : $\frac{∣ k + 1∣}{k ^{2} + 2 2} = \frac{3}{2} .$ 3. Squaring: $\frac{( k + 1 ) ^{2}}{2 ( k ^{2} + 2 )} = \frac{3}{4} \Rightarrow 4 (k + 1)^{2} = 6 (k^{2} + 2) \Rightarrow 2 k^{2} - 8 k + 8 = 0 \Rightarrow (k - 2)^{2} = 0$ . 4. $k = 2$ .

(ii) 1. $L_{3}$ : $(3 + 2 λ, λ, 1 - λ)$ . 2. Imposing $x + y = 12$ : $3 + 3 λ = 12$ , so $λ = 3$ . 3. $P = (9, 3, - 2)$ .

Answer: a) $r = (1, 0, 4) + t (1, 3, - 5)$ ; c) $x + 13 y + 8 z + 3 = 0$ ; d) (i) $k = 2$ ; (ii) $P (9, 3, - 2)$ . Part b) is proved.

Exercise 54 HardIB · P2 TZ2 May 2004

a) Consider the unit vectors $u$ and $v$ in three-dimensional space. Show that the vector $u + v$ bisects the angle between $u$ and $v$ .

Consider the points $A (2, 5, 4)$ , $B (1, 3, 2)$ , $C (5, 5, 6)$ . The line $L$ passes through $B$ and bisects the angle $A B C$ .

b) Find the equation of $L$ .

c) The line $L$ meets $[A C]$ at the point $D$ . Find the coordinates of $D$ .

Show answer

Answer: b) $r = 132 + t 11 4/3$ ; c) $D (3, 5, \frac{14}{3})$ . Part a) is thereby proved.

Show working and answer

a) 1. Let $φ$ be the angle between $u$ and $u + v$ , and $ψ$ the angle between $v$ and $u + v$ . As $∣ u ∣ = ∣ v ∣ = 1$ , $u \cdot (u + v) = ∣ u ∣^{2} + u \cdot v = 1 + u \cdot v, v \cdot (u + v) = u \cdot v + ∣ v ∣^{2} = 1 + u \cdot v .$ 2. Then $cos φ = \frac{u \cdot ( u + v )}{∣ u ∣ ∣ u + v ∣} = \frac{1 + u \cdot v}{∣ u + v ∣} = \frac{v \cdot ( u + v )}{∣ v ∣ ∣ u + v ∣} = cos ψ .$ 3. As $φ, ψ \in [0, π]$ and the cosine is one-to-one on that interval, $φ = ψ$ : the vector $u + v$ makes equal angles with $u$ and with $v$ , that is, it bisects the angle between them.

b) 1. The vectors from $B$ to $A$ and from $B$ to $C$ are $B A = 122, B C = 424, ∣ B A ∣ = 3, ∣ B C ∣ = 6.$ 2. Their unit vectors are $u = \frac{1}{3} 122, v = \frac{1}{6} 424 = \frac{1}{3} 212 .$ 3. By part a), the bisector has the direction of $u + v = 11 4/3$ , which is proportional to $334$ . 4. The line passes through $B (1, 3, 2)$ : $L : r = 132 + t 11 4/3, or x = 1 + t, y = 3 + t, z = 2 + \frac{4}{3} t .$

c) 1. A point of $[A C]$ is $A + s A C$ with $A C = 302$ and $0 \leq s \leq 1$ : $(2 + 3 s, 5, 4 + 2 s)$ . 2. Equating with $L$ : $1 + t = 2 + 3 s, 3 + t = 5, 2 + \frac{4}{3} t = 4 + 2 s$ . 3. From the second, $t = 2$ . From the first, $3 = 2 + 3 s$ , so $s = \frac{1}{3}$ . Checking the third: $2 + \frac{8}{3} = \frac{14}{3}$ and $4 + \frac{2}{3} = \frac{14}{3}$ . It holds. 4. Substituting $s = \frac{1}{3}$ : $D = (3, 5, \frac{14}{3})$ .

Answer: b) $r = 132 + t 11 4/3$ ; c) $D (3, 5, \frac{14}{3})$ . Part a) is thereby proved.

Exercise 55 HardIB · P1 TZ1 May 2021

Consider the line $L_{1}$ defined by the Cartesian equation $\frac{x + 1}{2} = y = 3 - z$ .

a) i) Show that the point $(- 1, 0, 3)$ belongs to $L_{1}$ .

(ii) Calculate the vector equation of $L_{1}$ .

Consider a second line $L_{2}$ defined by the vector equation

$r = 012 + t a 1 - 1$ where $t \in R$ and $a \in R$ .

b) Calculate the possible values of $a$ when the acute angle between $L_{1}$ and $L_{2}$ is $4 5^{\circ}$ .

The lines $L_{1}$ and $L_{2}$ have a unique point of intersection, $A$ , when $a \neq = k$ .

c) Calculate the value of $k$ and the coordinates of the point $A$ in terms of $a$ .

Show answer

Answer: a) ii) $r = - 1 03 + λ 21 - 1$ ; b) $a = - 4 \pm 32$ ; c) $k = 2$ and $A (\frac{a}{a - 2}, \frac{a - 1}{a - 2}, \frac{2 a - 5}{a - 2})$ .

Show working and answer

a) i) Substituting $(x, y, z) = (- 1, 0, 3)$ : $\frac{- 1 + 1}{2} = 0$ , $y = 0$ and $3 - z = 0$ . All three expressions equal $0$ , so the point belongs to $L_{1}$ .

(ii) 1. Call $λ$ the common value of the three expressions: $\frac{x + 1}{2} = λ$ , $y = λ$ , $3 - z = λ$ . Then $x = - 1 + 2 λ$ , $y = λ$ , $z = 3 - λ$ . 2. In vector form: $L_{1} : r = - 1 03 + λ 21 - 1 .$

b) 1. The direction vectors are $d_{1} = (2, 1, - 1)$ and $d_{2} = (a, 1, - 1)$ . The acute angle $θ$ satisfies $cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$ . 2. $d_{1} \cdot d_{2} = 2 a + 1 + 1 = 2 a + 2$ , $∣ d_{1} ∣ = 6$ and $∣ d_{2} ∣ = a^{2} + 2$ . 3. With $θ = 4 5^{\circ}$ : $\frac{∣2 a + 2∣}{6 a ^{2} + 2} = \frac{2}{2} \Rightarrow \frac{4 ( a + 1 ) ^{2}}{6 ( a ^{2} + 2 )} = \frac{1}{2} \Rightarrow 4 (a + 1)^{2} = 3 (a^{2} + 2) .$ 4. Expanding: $4 a^{2} + 8 a + 4 = 3 a^{2} + 6$ , that is, $a^{2} + 8 a - 2 = 0$ . 5. $a = \frac{- 8 \pm 64 + 8}{2} = - 4 \pm 32$ .

c) 1. A common point satisfies: $- 1 + 2 λ = a t$ , $λ = 1 + t$ , $3 - λ = 2 - t$ . 2. The last two equations are equivalent ( $λ = 1 + t$ ). Substituting into the first: $- 1 + 2 (1 + t) = a t$ , that is, $1 + 2 t = a t$ and $(a - 2) t = 1$ . 3. This equation has a unique solution $t = \frac{1}{a - 2}$ if and only if $a \neq = 2$ . Therefore $k = 2$ (for $a = 2$ the lines are parallel and do not intersect). 4. Coordinates of $A$ , substituting $t = \frac{1}{a - 2}$ into $L_{2}$ : $x = a t = \frac{a}{a - 2}, y = 1 + t = \frac{a - 1}{a - 2}, z = 2 - t = \frac{2 a - 5}{a - 2} .$

Answer: a) ii) $r = - 1 03 + λ 21 - 1$ ; b) $a = - 4 \pm 32$ ; c) $k = 2$ and $A (\frac{a}{a - 2}, \frac{a - 1}{a - 2}, \frac{2 a - 5}{a - 2})$ .

Exercise 56 HardIB · P2 TZ0 November 2022

Consider the points $A (1, 2, 3)$ , $B (k, - 2, 1)$ and $C (5, 0, 2)$ , where $k \in R$ .

a) Write down $A B$ and $A C$ .

b) Given that the points $A$ , $B$ and $C$ lie on a straight line, show that $k = 9$ .

c) For $k = 9$ , let $L_{1}$ be the line passing through $A$ , $B$ and $C$ .

(i) Find the vector equation of the line $L_{1}$ .

(ii) The line $L_{2}$ has equation $\frac{x - 1}{2} = \frac{y}{3} = 1 - z$ .

Show that the lines $L_{1}$ and $L_{2}$ are skew.

d) For $k \neq = 9$ , let $Π$ be the plane containing $A$ , $B$ and $C$ .

(i) Find the Cartesian equation of the plane $Π$ .

(ii) Find the coordinates of the point on the plane $Π$ that is closest to the origin $(0, 0, 0)$ .

Show answer

Answer: a) $A B = k - 1 - 4 - 2$ , $A C = 4 - 2 - 1$ ; c) i) $r = 123 + λ 4 - 2 - 1$ ; d) i) $Π : y - 2 z = - 4$ ; ii) $(0, - \frac{4}{5}, \frac{8}{5})$ . Parts b) and c) ii) are shown.

Show working and answer

a) $A B = k - 1 - 4 - 2$ and $A C = 4 - 2 - 1$ .

b) 1. Three points are collinear when $A B$ and $A C$ are parallel: $A B = m A C$ . 2. From the second component: $- 4 = - 2 m$ , so $m = 2$ . The third component confirms it: $- 2 = 2 \cdot (- 1)$ . 3. From the first component: $k - 1 = 2 \cdot 4 = 8$ , so $k = 9$ .

c) (i) 1. With $k = 9$ the line passes through $A (1, 2, 3)$ and has direction $A C = (4, - 2, - 1)$ : $L_{1} : r = 123 + λ 4 - 2 - 1 .$

(ii) 1. Write $L_{2}$ in parametric form, with $\frac{x - 1}{2} = \frac{y}{3} = 1 - z = μ$ : $x = 1 + 2 μ$ , $y = 3 μ$ , $z = 1 - μ$ . It passes through $(1, 0, 1)$ with direction $(2, 3, - 1)$ . 2. The directions $(4, - 2, - 1)$ and $(2, 3, - 1)$ are not parallel (they are not proportional), so the lines are neither parallel nor coincident. 3. Check whether they intersect: $1 + 4 λ = 1 + 2 μ$ , $2 - 2 λ = 3 μ$ and $3 - λ = 1 - μ$ . From the first, $μ = 2 λ$ . Substituting into the second: $2 - 2 λ = 6 λ$ , so $λ = \frac{1}{4}$ and $μ = \frac{1}{2}$ . 4. Third equation: $3 - \frac{1}{4} = \frac{11}{4}$ compared with $1 - \frac{1}{2} = \frac{1}{2}$ . It is not satisfied, so there is no common point. 5. The lines are not parallel and do not intersect: they are skew.

d) (i) 1. A normal to the plane is $A B \times A C = (- 4) (- 1) - (- 2) (- 2) (- 2) (4) - (k - 1) (- 1) (k - 1) (- 2) - (- 4) (4) = 0 k - 9 18 - 2 k = (k - 9) 01 - 2,$ which is not zero because $k \neq = 9$ . A simple normal is $(0, 1, - 2)$ . 2. The plane is $y - 2 z = d$ ; with $A$ : $d = 2 - 6 = - 4$ . 3. $Π : y - 2 z = - 4$ . Check: $B$ : $- 2 - 2 = - 4$ and $C$ : $0 - 4 = - 4$ .

(ii) 1. The point of $Π$ closest to the origin is the foot of the perpendicular from the origin: the line $(0, s, - 2 s)$ (the direction of the normal). 2. Substitute into the plane: $s - 2 (- 2 s) = 5 s = - 4$ , so $s = - \frac{4}{5}$ . 3. The point is $(0, - \frac{4}{5}, \frac{8}{5})$ (at distance $\frac{4}{5}$ from the origin).

Answer: a) $A B = k - 1 - 4 - 2$ , $A C = 4 - 2 - 1$ ; c) i) $r = 123 + λ 4 - 2 - 1$ ; d) i) $Π : y - 2 z = - 4$ ; ii) $(0, - \frac{4}{5}, \frac{8}{5})$ . Parts b) and c) ii) are shown.

Equations of a plane

Exercise 57 BasicIB · P1 November 1999

The vector $n = 2 i - j + 3 k$ is normal to the plane that passes through the point $(2, 1, 2)$ .

a) Calculate an equation of the plane.

b) Calculate $a$ if the point $(a, a - 1, a - 2)$ lies on the plane.

Show answer

Answer: a) $2 x - y + 3 z - 9 = 0$ ; b) $a = \frac{7}{2}$

Show working and answer

a) 1. A plane with normal vector $(2, - 1, 3)$ has equation $2 x - y + 3 z = d$ . 2. Substituting $(2, 1, 2)$ : $d = 4 - 1 + 6 = 9$ , that is, $2 x - y + 3 z - 9 = 0$ .

b) 1. Substituting $(a, a - 1, a - 2)$ into the equation: $2 a - (a - 1) + 3 (a - 2) = 9$ . 2. Expanding: $2 a - a + 1 + 3 a - 6 = 9$ , that is, $4 a - 5 = 9$ , so $a = \frac{14}{4} = \frac{7}{2}$ .

Answer: a) $2 x - y + 3 z - 9 = 0$ ; b) $a = \frac{7}{2}$

Exercise 58 BasicIB · P1 May 2000

Calculate a normal vector to the plane that contains the lines $L_{1}$ and $L_{2}$ , whose equations are:

$L_{1} : r = i + k + λ (2 i + j - 2 k)$

$L_{2} : r = 3 i + 2 j + 2 k + μ (j + 3 k)$

Show answer

Answer: $n = 5 i - 6 j + 2 k$ (or any multiple of it)

Show working and answer

The direction vectors of the lines lie in the plane: $d_{1} = 21 - 2$ and $d_{2} = 013$ .
A normal vector is perpendicular to both, so we obtain it with the cross product: $n = d_{1} \times d_{2} = 1 \cdot 3 - (- 2) \cdot 1 (- 2) \cdot 0 - 2 \cdot 3 2 \cdot 1 - 1 \cdot 0 = 5 - 6 2 .$

Answer: $n = 5 i - 6 j + 2 k$ (or any multiple of it)

Exercise 59 BasicIB · P1 May 2000

The plane $6 x - 2 y + z = 11$ contains the line $x - 1 = \frac{y + 1}{2} = \frac{z - 3}{l}$ . Calculate $l$ .

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Answer: $l = - 2$

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The line passes through $(1, - 1, 3)$ and has direction vector $d = 12 l$ .
If the line lies in the plane, its direction vector is perpendicular to the normal vector of the plane $n = 6 - 2 1$ : $d \cdot n = 0$ .
$6 - 4 + l = 0$ , so $l = - 2$ .
Check that the point $(1, - 1, 3)$ lies on the plane: $6 + 2 + 3 = 11$ . Correct.

Answer: $l = - 2$

Exercise 60 MediumIB · P2 TZ0 November 2009

The vector equation of the line $L$ is given by $x y z = 136 + λ - 1 2 - 1 .$

Find the Cartesian equation of the plane that contains the line $L$ and the point $A (4, - 2, 5)$ .

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Answer: $7 x + 4 y + z = 25$

Show working and answer

A point on $L$ is $P (1, 3, 6)$ , and its direction vector is $d = - 1 2 - 1$ .
A second vector in the plane is $P A = 3 - 5 - 1$ .
A normal vector to the plane is $n = d \times P A = 2 \cdot (- 1) - (- 1) (- 5) (- 1) \cdot 3 - (- 1) (- 1) (- 1) (- 5) - 2 \cdot 3 = - 7 - 4 - 1 .$ We use the opposite vector $741$ , which is also normal.
The plane is $7 x + 4 y + z = d$ . Substituting $A (4, - 2, 5)$ : $d = 28 - 8 + 5 = 25$ .
Check with $P$ : $7 + 12 + 6 = 25$ .

Answer: $7 x + 4 y + z = 25$

Exercise 61 MediumIB · P1 May 2001

Find an equation of the plane that contains the two lines $x - 1 = \frac{1 - y}{2} = z - 2 and \frac{x + 1}{3} = \frac{2 - y}{3} = \frac{z + 2}{5} .$

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Answer: $- 7 x - 2 y + 3 z = - 3$ , that is, $7 x + 2 y - 3 z = 3$

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First line: calling $t$ the common value, $x = 1 + t$ , $y = 1 - 2 t$ , $z = 2 + t$ . It passes through $A (1, 1, 2)$ with $d_{1} = 1 - 2 1$ .
Second line: $x = - 1 + 3 s$ , $y = 2 - 3 s$ , $z = - 2 + 5 s$ . It passes through $B (- 1, 2, - 2)$ with $d_{2} = 3 - 3 5$ .
A normal vector to the plane is $n = d_{1} \times d_{2} = (- 2) \cdot 5 - 1 \cdot (- 3) 1 \cdot 3 - 1 \cdot 5 1 \cdot (- 3) - (- 2) \cdot 3 = - 7 - 2 3 .$
The plane is $- 7 x - 2 y + 3 z = d$ . With $A (1, 1, 2)$ : $d = - 7 - 2 + 6 = - 3$ .
Check with $B (- 1, 2, - 2)$ : $7 - 4 - 6 = - 3$ . Correct, so the lines lie in the same plane.

Answer: $- 7 x - 2 y + 3 z = - 3$ , that is, $7 x + 2 y - 3 z = 3$

Exercise 62 MediumIB · P1 November 1998

Consider the planes $P$ and $Q$ with equations: $P : 6 x - 2 y + 3 z + 8 = 0$ , $Q : x - 3 y + 5 z = 0$ . Find the equation of a third plane $R$ , which is perpendicular to $P$ and $Q$ and contains the point $(2, 2, 3)$ .

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Answer: $R : x + 27 y + 16 z - 104 = 0$

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The normal vectors are $n_{P} = (6, - 2, 3)$ and $n_{Q} = (1, - 3, 5)$ .
A plane perpendicular to $P$ and $Q$ has a normal that is perpendicular to both $n_{P}$ and $n_{Q}$ , that is, parallel to their vector product: $n_{P} \times n_{Q} = (- 2) \cdot 5 - 3 \cdot (- 3) 3 \cdot 1 - 6 \cdot 5 6 \cdot (- 3) - (- 2) \cdot 1 = - 1 - 27 - 16 ∥ 12716 .$
The plane is $x + 27 y + 16 z = d$ . With $(2, 2, 3)$ : $d = 2 + 54 + 48 = 104$ .

Answer: $R : x + 27 y + 16 z - 104 = 0$

Exercise 63 MediumIB · P2 TZ0 November 2014

The lines $L_{1}$ and $L_{2}$ are given by $L_{1} : \frac{x - 1}{3} = \frac{y - 5}{2} = \frac{z - 12}{- 2}$ , $L_{2} : \frac{x - 1}{8} = \frac{y - 5}{11} = \frac{z - 12}{6}$ .

The plane $π$ contains $L_{1}$ and also $L_{2}$ .

a) Find the Cartesian equation of $π$ .

The line $L_{3}$ passes through the point $(4, 0, 8)$ and is perpendicular to $π$ .

b) Find the coordinates of the point at which $L_{3}$ meets $π$ .

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Answer: a) $2 x - 2 y + z - 4 = 0$ ; b) $(\frac{4}{3}, \frac{8}{3}, \frac{20}{3})$

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a) 1. Both lines pass through $(1, 5, 12)$ and have direction vectors $(3, 2, - 2)$ and $(8, 11, 6)$ . 2. A normal vector to the plane is $32 - 2 \times 8116 = 2 \cdot 6 - (- 2) \cdot 11 (- 2) \cdot 8 - 3 \cdot 6 3 \cdot 11 - 2 \cdot 8 = 34 - 34 17 ∥ 2 - 2 1 .$ 3. The plane is $2 x - 2 y + z = d$ . With $(1, 5, 12)$ : $d = 2 - 10 + 12 = 4$ .

b) 1. The line $L_{3}$ has as direction vector the normal $(2, - 2, 1)$ : $(4 + 2 t, - 2 t, 8 + t)$ . 2. We substitute into the plane: $2 (4 + 2 t) - 2 (- 2 t) + (8 + t) = 4$ , that is, $9 t + 16 = 4$ , so $t = - \frac{4}{3}$ . 3. The point is $(4 - \frac{8}{3}, \frac{8}{3}, 8 - \frac{4}{3}) = (\frac{4}{3}, \frac{8}{3}, \frac{20}{3})$ .

Answer: a) $2 x - 2 y + z - 4 = 0$ ; b) $(\frac{4}{3}, \frac{8}{3}, \frac{20}{3})$

Exercise 64 MediumIB · P2 TZ1 May 2017

Find the Cartesian equation of the plane $Π$ that contains the points $A (6, 2, 1)$ and $B (3, - 1, 1)$ and is perpendicular to the plane $x + 2 y - z - 6 = 0$ .

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Answer: $Π : x - y - z - 3 = 0$

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The required plane contains the direction $A B = (- 3, - 3, 0)$ .
If it is perpendicular to the given plane, its normal is perpendicular to the normal $n_{0} = (1, 2, - 1)$ of that plane; that is, $n_{0}$ is parallel to the required plane.
Normal of the required plane: $A B \times n_{0} = (- 3) (- 1) - 0 \cdot 2 0 \cdot 1 - (- 3) (- 1) (- 3) \cdot 2 - (- 3) \cdot 1 = 3 - 3 - 3$ ; we use $n = (1, - 1, - 1)$ .
$x - y - z = d$ with $A (6, 2, 1)$ : $d = 6 - 2 - 1 = 3$ . We check with $B$ : $3 + 1 - 1 = 3$ .

Answer: $Π : x - y - z - 3 = 0$

Exercise 65 MediumIB · P1 TZ0 November 2016

Consider the vectors $a = i - 3 j - 2 k$ , $b = - 3 j + 2 k$ .

a) Find $a \times b$ .

b) Hence, find the Cartesian equation of the plane that contains the vectors $a$ and $b$ , and passes through the point $(1, 0, - 1)$ .

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Answer: a) $a \times b = - 12 i - 2 j - 3 k$ ; b) $12 x + 2 y + 3 z = 9$ (equivalent to $- 12 x - 2 y - 3 z + 9 = 0$ )

Show working and answer

a) 1. With the components $a = (1, - 3, - 2)$ and $b = (0, - 3, 2)$ : $a \times b = (- 3) (2) - (- 2) (- 3) (- 2) (0) - (1) (2) (1) (- 3) - (- 3) (0) = - 12 - 2 - 3 .$ That is, $a \times b = - 12 i - 2 j - 3 k$ .

b) 1. The cross product is perpendicular to $a$ and to $b$ , so it is a normal vector of the plane: $n = (- 12, - 2, - 3)$ (or its opposite $(12, 2, 3)$ ). 2. The equation is $12 x + 2 y + 3 z = d$ . Imposing that it passes through $(1, 0, - 1)$ : $d = 12 + 0 - 3 = 9$ . 3. The plane is $12 x + 2 y + 3 z = 9$ , or equivalently, $- 12 x - 2 y - 3 z + 9 = 0$ .

Answer: a) $a \times b = - 12 i - 2 j - 3 k$ ; b) $12 x + 2 y + 3 z = 9$ (equivalent to $- 12 x - 2 y - 3 z + 9 = 0$ )

Exercise 66 MediumIB · P1 November 1998

Consider the two planes $P$ and $Q$ , where $P : 6 x - 2 y + 3 z + 8 = 0$ and $Q : x - 3 y + 5 = 0$ .

Find the equation of a third plane $R$ which is perpendicular to $P$ and $Q$ and contains the point $(2, 2, 3)$ .

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Answer: $R : 9 x + 3 y - 16 z + 24 = 0$

Show working and answer

The normals of $P$ and $Q$ are $n_{P} = (6, - 2, 3)$ and $n_{Q} = (1, - 3, 0)$ .
A plane is perpendicular to $P$ and to $Q$ when its normal is perpendicular to $n_{P}$ and to $n_{Q}$ . We take $n_{R} = n_{P} \times n_{Q} = (- 2) \cdot 0 - 3 (- 3) 3 \cdot 1 - 6 \cdot 0 6 (- 3) - (- 2) \cdot 1 = 93 - 16 .$
The plane is $9 x + 3 y - 16 z = d$ . As it contains $(2, 2, 3)$ : $d = 9 \cdot 2 + 3 \cdot 2 - 16 \cdot 3 = 18 + 6 - 48 = - 24$ .
Thus $R : 9 x + 3 y - 16 z = - 24$ .

Answer: $R : 9 x + 3 y - 16 z + 24 = 0$

Exercise 67 MediumIB · P1 TZ1 May 2010

a) Show that the two planes $π_{1} : x + 2 y - z = 1$ and $π_{2} : x + z = - 2$ are perpendicular.

b) Find the equation of the plane $π_{3}$ that passes through the origin and is perpendicular to $π_{1}$ and $π_{2}$ .

Show answer

Answer: a) $n_{1} \cdot n_{2} = 0$ , so they are perpendicular. b) $π_{3} : x - y - z = 0$ . This completes the proof of part a).

Show working and answer

a) 1. The normal vectors are $n_{1} = 12 - 1$ and $n_{2} = 101$ . 2. Two planes are perpendicular if their normals are perpendicular, that is, if $n_{1} \cdot n_{2} = 0$ : $n_{1} \cdot n_{2} = 1 \cdot 1 + 2 \cdot 0 + (- 1) \cdot 1 = 0.$ 3. Therefore $π_{1} ⊥ π_{2}$ .

b) 1. A plane perpendicular to $π_{1}$ and $π_{2}$ has a normal vector perpendicular to $n_{1}$ and $n_{2}$ . The vector (cross) product works: $n_{1} \times n_{2} = 2 \cdot 1 - (- 1) \cdot 0 (- 1) \cdot 1 - 1 \cdot 1 1 \cdot 0 - 2 \cdot 1 = 2 - 2 - 2 .$ 2. We can use the proportional vector $n_{3} = 1 - 1 - 1$ , so $π_{3} : x - y - z = d$ . 3. It passes through the origin, so $d = 0$ and $π_{3} : x - y - z = 0$ .

Answer: a) $n_{1} \cdot n_{2} = 0$ , so they are perpendicular. b) $π_{3} : x - y - z = 0$ . This completes the proof of part a).

Exercise 68 MediumIB · P1 TZ1 May 2021

Consider the planes $Π_{1}$ and $Π_{2}$ with the following equations:

$Π_{1} : 3 x + 2 y + z = 6$

$Π_{2} : x - 2 y + z = 4$

a) Calculate the Cartesian equation of the plane $Π_{3}$ which is perpendicular to $Π_{1}$ and $Π_{2}$ and passes through the origin $(0, 0, 0)$ .

b) Calculate the coordinates of the point where $Π_{1}$ , $Π_{2}$ and $Π_{3}$ intersect.

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Answer: a) $Π_{3} : 2 x - y - 4 z = 0$ ; b) $(\frac{41}{21}, - \frac{10}{21}, \frac{23}{21}) \approx (1.95; - 0.476; 1.10)$ .

Show working and answer

a) 1. A plane perpendicular to $Π_{1}$ and $Π_{2}$ has as normal a vector perpendicular to the normals $n_{1} = (3, 2, 1)$ and $n_{2} = (1, - 2, 1)$ (equivalently, its normal is parallel to the line of intersection of both): $n_{3} = n_{1} \times n_{2} = 2 \cdot 1 - 1 \cdot (- 2) 1 \cdot 1 - 3 \cdot 1 3 (- 2) - 2 \cdot 1 = 4 - 2 - 8 = 2 2 - 1 - 4 .$ 2. The plane is $2 x - y - 4 z = d$ and passes through the origin, so $d = 0$ : $Π_{3} : 2 x - y - 4 z = 0.$

b) 1. Adding the equations of $Π_{1}$ and $Π_{2}$ : $4 x + 2 z = 10$ , so $z = 5 - 2 x$ . 2. Subtracting them: $2 x + 4 y = 2$ , so $y = \frac{1 - x}{2}$ . 3. Substituting into $Π_{3}$ : $2 x - \frac{1 - x}{2} - 4 (5 - 2 x) = 0$ . Multiplying by $2$ : $4 x - 1 + x - 40 + 16 x = 0$ , that is, $21 x = 41$ and $x = \frac{41}{21}$ . 4. Then $y = \frac{1 - \frac{41}{21}}{2} = - \frac{10}{21}$ and $z = 5 - \frac{82}{21} = \frac{23}{21}$ .

Answer: a) $Π_{3} : 2 x - y - 4 z = 0$ ; b) $(\frac{41}{21}, - \frac{10}{21}, \frac{23}{21}) \approx (1.95; - 0.476; 1.10)$ .

Exercise 69 HardIB · P2 May 1999

The coordinates of the points $P$ , $Q$ , $R$ and $S$ are $(4, 1, - 1)$ , $(3, 3, 5)$ , $(1, 0, 2 c)$ and $(1, 1, 2)$ respectively.

a) Calculate the value of $c$ for which the vectors $QR$ and $P R$ are perpendicular.

For the rest of the question, use the value of $c$ found in part (a) for the coordinate of point $R$ .

b) Find $P S \times P R$ .

c) Calculate the equation of the line $l$ that passes through the point $Q$ and is parallel to the vector $P R$ .

d) Calculate the equation of the plane $π$ that contains the line $l$ and passes through the point $S$ .

e) Calculate the shortest distance from the point $P$ to the plane $π$ .

Show answer

Answer: a) $c = 1$ ; b) $3 i + 3 k$ ; c) $l : \frac{x - 3}{- 3} = \frac{y - 3}{- 1} = \frac{z - 5}{3}$ ; d) $π : 9 x - 15 y + 4 z = 2$ ; e) $\frac{15}{322} \approx 0.836$

Show working and answer

a) 1. $QR = - 2 - 3 2 c - 5$ and $P R = - 3 - 1 2 c + 1$ . 2. Perpendicular vectors have zero scalar product: $6 + 3 + (2 c - 5) (2 c + 1) = 0$ . 3. Expanding: $9 + 4 c^{2} - 8 c - 5 = 0$ , that is, $4 c^{2} - 8 c + 4 = 4 (c - 1)^{2} = 0$ , so $c = 1$ .

With $c = 1$ : $R (1, 0, 2)$ , $P R = - 3 - 1 3$ .

b) 1. $P S = - 3 03$ . 2. $P S \times P R = 0 \cdot 3 - 3 \cdot (- 1) 3 \cdot (- 3) - (- 3) \cdot 3 (- 3) (- 1) - 0 \cdot (- 3) = 303 .$

c) 1. The line passes through $Q (3, 3, 5)$ with direction vector $P R = - 3 - 1 3$ . 2. In symmetric form: $l : \frac{x - 3}{- 3} = \frac{y - 3}{- 1} = \frac{z - 5}{3}$ .

d) 1. The plane contains $l$ , so it contains the vector $P R$ (the direction vector of $l$ ). It also contains $QS = - 2 - 2 - 3$ , which joins $Q \in l$ to $S$ . 2. A normal vector is $n = P R \times QS = (- 1) (- 3) - 3 \cdot (- 2) 3 \cdot (- 2) - (- 3) (- 3) (- 3) (- 2) - (- 1) (- 2) = 9 - 15 4 .$ 3. The plane is $9 x - 15 y + 4 z = d$ . With $Q (3, 3, 5)$ : $d = 27 - 45 + 20 = 2$ . With $S (1, 1, 2)$ as well: $9 - 15 + 8 = 2$ .

e) 1. Distance from $P (4, 1, - 1)$ to the plane $9 x - 15 y + 4 z - 2 = 0$ : $d = \frac{∣9 \cdot 4 - 15 \cdot 1 + 4 \cdot ( - 1 ) - 2∣}{9 ^{2} + 1 5 ^{2} + 4 ^{2}} = \frac{∣36 - 15 - 4 - 2∣}{322} = \frac{15}{322} \approx 0.836.$

Answer: a) $c = 1$ ; b) $3 i + 3 k$ ; c) $l : \frac{x - 3}{- 3} = \frac{y - 3}{- 1} = \frac{z - 5}{3}$ ; d) $π : 9 x - 15 y + 4 z = 2$ ; e) $\frac{15}{322} \approx 0.836$

Exercise 70 HardIB · P2 May 2002

The points $A$ , $B$ , $C$ and $D$ have the following coordinates:

$A (1, 3, 1)$ , $B (1, 2, 4)$ , $C (2, 3, 6)$ , $D (5, - 2, 1)$

a) i) Calculate the vector product $A B \times A C$ , expressing your answer in terms of the unit vectors $i$ , $j$ , $k$ .

(ii) Find the area of triangle $A B C$ .

Let $Π$ be the plane containing the points $A$ , $B$ and $C$ , and let $L$ be the line through $D$ which is perpendicular to $Π$ .

The point of intersection of $L$ and $Π$ is $P$ .

b) i) Find the Cartesian equation of $Π$ .

(ii) Find the Cartesian equation of $L$ .

c) Find the coordinates of $P$ .

d) Find the perpendicular distance from $D$ to $Π$ .

Show answer

Answer: a) i) $- 5 i + 3 j + k$ ; ii) $\frac{35}{2}$ ; b) i) $Π : - 5 x + 3 y + z - 5 = 0$ ; ii) $L : \frac{x - 5}{- 5} = \frac{y + 2}{3} = \frac{z - 1}{1}$ ; c) $P (0, 1, 2)$ ; d) $35$

Show working and answer

a) i) 1. $A B = 0 - 1 3$ and $A C = 105$ . 2. $A B \times A C = (- 1) \cdot 5 - 3 \cdot 0 3 \cdot 1 - 0 \cdot 5 0 \cdot 0 - (- 1) \cdot 1 = - 5 31 = - 5 i + 3 j + k .$

(ii) The area is $\frac{1}{2} A B \times A C = \frac{1}{2} 25 + 9 + 1 = \frac{35}{2}$ .

b) i) 1. The vector $A B \times A C = (- 5, 3, 1)$ is normal to $Π$ , so $Π$ has the form $- 5 x + 3 y + z = d$ . 2. With $A (1, 3, 1)$ : $d = - 5 + 9 + 1 = 5$ . We check with $B$ : $- 5 + 6 + 4 = 5$ , and with $C$ : $- 10 + 9 + 6 = 5$ . 3. $Π : - 5 x + 3 y + z - 5 = 0$ .

(ii) 1. $L$ passes through $D (5, - 2, 1)$ and has as direction vector the normal of $Π$ , $(- 5, 3, 1)$ . 2. $L : \frac{x - 5}{- 5} = \frac{y + 2}{3} = \frac{z - 1}{1}$ .

c) 1. Points of $L$ : $(5 - 5 t, - 2 + 3 t, 1 + t)$ . 2. We substitute into $Π$ : $- 5 (5 - 5 t) + 3 (- 2 + 3 t) + (1 + t) = 5$ , that is, $- 25 + 25 t - 6 + 9 t + 1 + t = 5$ , so $35 t = 35$ and $t = 1$ . 3. $P (0, 1, 2)$ .

d) The distance from $D$ to $Π$ is $∣ D P ∣$ with $D P = (- 5, 3, 1)$ : $25 + 9 + 1 = 35$ . (Using the formula: $\frac{∣ - 25 - 6 + 1 - 5∣}{35} = \frac{35}{35} = 35$ .)

Answer: a) i) $- 5 i + 3 j + k$ ; ii) $\frac{35}{2}$ ; b) i) $Π : - 5 x + 3 y + z - 5 = 0$ ; ii) $L : \frac{x - 5}{- 5} = \frac{y + 2}{3} = \frac{z - 1}{1}$ ; c) $P (0, 1, 2)$ ; d) $35$

Exercise 71 HardIB · P1 TZ2 May 2009

a) Verify that a Cartesian equation of the line $L_{1}$ , which contains the points $A (1, - 1, 2)$ and $B (3, 0, 3)$ , is of the form $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 2}{1}$ .

b) A second line $L_{2}$ has an equation of the form $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{1}$ .

Verify that the lines $L_{1}$ and $L_{2}$ intersect, and find the coordinates of the point of intersection.

c) Given that $d_{1}$ and $d_{2}$ are direction vectors of $L_{1}$ and $L_{2}$ respectively, determine $d_{1} \times d_{2}$ .

d) Verify that a Cartesian equation of the plane $π$ , which contains $L_{1}$ and $L_{2}$ , is $- x - y + 3 z = 6$ .

e) Find a vector equation of the line $L_{3}$ , which is perpendicular to the plane $π$ and passes through the point $T (3, 1, - 4)$ .

f) i) Find the point of intersection of the line $L_{3}$ and the plane $π$ .

(ii) Find the coordinates of $T^{'}$ , the reflection of $T$ in the plane $π$ .

(iii) Hence, find the modulus of the vector $T T^{'}$ .

Show answer

Answer: b) $(- 1, - 2, 1)$ ; c) $- i - j + 3 k$ ; e) $r = 31 - 4 + t - 1 - 1 3$ ; f) i) $(1, - 1, 2)$ ; ii) $T^{'} (- 1, - 3, 8)$ ; iii) $∣ T T^{'} ∣ = 411$ . Parts a) and d) are verified.

Show working and answer

a) 1. $A B = (2, 1, 1)$ is a direction vector of $L_{1}$ , which passes through $A (1, - 1, 2)$ : the symmetric equation is $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 2}{1}$ . 2. We check with $B$ : $\frac{3 - 1}{2} = 1$ , $\frac{0 + 1}{1} = 1$ , $\frac{3 - 2}{1} = 1$ . All three ratios equal $1$ , so $B$ lies on the line.

b) 1. Parametric form: $L_{1} : (1 + 2 s, - 1 + s, 2 + s)$ and $L_{2} : (1 + μ, 2 + 2 μ, 3 + μ)$ . 2. From the first coordinate, $μ = 2 s$ . From the second: $- 1 + s = 2 + 4 s$ , so $s = - 1$ and $μ = - 2$ . 3. Third coordinate: $2 + (- 1) = 1$ and $3 + (- 2) = 1$ . They coincide, so the lines intersect. 4. Point: $(1 - 2, - 1 - 1, 2 - 1) = (- 1, - 2, 1)$ .

c) 1. With $d_{1} = (2, 1, 1)$ and $d_{2} = (1, 2, 1)$ : $d_{1} \times d_{2} = 1 \cdot 1 - 1 \cdot 2 1 \cdot 1 - 2 \cdot 1 2 \cdot 2 - 1 \cdot 1 = - 1 - 1 3 = - i - j + 3 k .$

d) 1. The vector $(- 1, - 1, 3)$ is normal to the plane containing both lines: $- x - y + 3 z = d$ . 2. With $A (1, - 1, 2)$ : $d = - 1 + 1 + 6 = 6$ . So $π : - x - y + 3 z = 6$ .

e) 1. The line $L_{3}$ has as direction vector the normal $(- 1, - 1, 3)$ and passes through $T$ : $r = 31 - 4 + t - 1 - 1 3 .$

f) i) 1. We substitute $(3 - t, 1 - t, - 4 + 3 t)$ into the plane: $- (3 - t) - (1 - t) + 3 (- 4 + 3 t) = 6$ , that is, $11 t - 16 = 6$ , so $t = 2$ . 2. The point of intersection is $(1, - 1, 2)$ .

f) ii) 1. The point of intersection is the midpoint of $T$ and $T^{'}$ : $T^{'} = 2 (1, - 1, 2) - (3, 1, - 4) = (- 1, - 3, 8)$ .

f) iii) 1. $T T^{'} = (- 4, - 4, 12)$ and $∣ T T^{'} ∣ = 16 + 16 + 144 = 176 = 411$ .

Exercise 72 HardIB · P1 TZ1 May 2014

a) Show that the points $O (0, 0, 0)$ , $A (6, 0, 0)$ , $B (6, - 24, 12)$ , $C (0, - 24, 12)$ form a square.

b) Calculate the coordinates of $M$ , the midpoint of $[OB]$ .

c) Show that an equation of the plane $Π$ , which contains the square $OABC$ , is $y + 2 z = 0$ .

d) Find a vector equation of the line $L$ , which passes through $M$ and is perpendicular to the plane $Π$ .

e) Find the coordinates of $D$ , the point of intersection of the line $L$ with the plane $y = 0$ .

f) Find the coordinates of $E$ , the reflection of the point $D$ in the plane $Π$ .

g) (i) Calculate the angle $ODA$ .

(ii) State what this tells you about the solid $OABCDE$ .

Show answer

Answer: b) $M (3, - 6, 3)$ ; d) $r = (3, - 6, 3) + t (0, 1, 2)$ ; e) $D (3, 0, 33)$ ; f) $E (3, - 26, - 3)$ ; g) (i) $6 0^{\circ}$ ; (ii) $O A B C D E$ is a regular octahedron. Parts a) and c) are proved.

Show working and answer

a) 1. We calculate the vectors of the sides: $O A = 600$ , $O C = 0 - 24 12$ , $A B = B - A = 0 - 24 12$ , $C B = B - C = 600$ . 2. As $A B = O C$ and $C B = O A$ , the quadrilateral $O A B C$ is a parallelogram. 3. Lengths of two consecutive sides: $∣ O A ∣ = 6$ and $∣ O C ∣ = 24 + 12 = 6$ . They are equal, so all four sides measure $6$ . 4. Dot product: $O A \cdot O C = 6 \cdot 0 + 0 + 0 = 0$ , so $O A ⊥ O C$ and there is a right angle. 5. A parallelogram with all four sides equal and a right angle is a square.

b) $M = \frac{O + B}{2} = (3, - \frac{24}{2}, \frac{12}{2}) = (3, - 6, 3)$ , since $24 = 26$ and $12 = 23$ .

c) 1. A normal vector to the plane is $O A \times O C = 600 \times 0 - 24 12 = 0 - 612 - 624 = - 612012$ , because $24 = 212$ . 2. We take $n = 012$ . The plane passes through $O$ , so its equation is $y + 2 z = 0$ . 3. Check: $A$ satisfies $0 + 0 = 0$ ; $C$ : $- 24 + 212 = 0$ ; $B$ : the same. This proves the result.

d) The line $L$ has as direction vector the normal $n$ and passes through $M$ : $r = 3 - 6 3 + t 012 .$

e) 1. A point of $L$ is $(3, - 6 + t, 3 + 2 t)$ . 2. We impose $y = 0$ : $t = 6$ . 3. Then $z = 3 + 26 = 3 + 12 = 3 + 23 = 33$ . Therefore $D = (3, 0, 33)$ .

f) 1. $M$ lies in $Π$ (it is the midpoint of $[O B]$ , and $O$ and $B$ lie in $Π$ ) and $L$ is perpendicular to $Π$ through $M$ . Therefore $M$ is the midpoint of $[D E]$ . 2. $E = 2 M - D = (6 - 3, - 26 - 0, 23 - 33) = (3, - 26, - 3)$ .

g) (i) 1. $D O = - 3 0 - 33$ and $D A = 30 - 33$ , with $∣ D O ∣ = ∣ D A ∣ = 9 + 27 = 6$ . 2. $cos O D A = \frac{D O \cdot D A}{6 \cdot 6} = \frac{- 9 + 27}{36} = \frac{1}{2}$ , so $O D A = 6 0^{\circ}$ .

(ii) Triangle $O D A$ has $∣ O D ∣ = ∣ D A ∣ = ∣ O A ∣ = 6$ : it is equilateral. In the same way $∣ D B ∣ = ∣ D C ∣ = 6$ , and $E$ , the reflection of $D$ in $Π$ , is also at distance $6$ from $O$ , $A$ , $B$ and $C$ . All twelve edges of the solid $O A B C D E$ measure $6$ and all its faces are equilateral triangles: it is a regular octahedron.

Exercise 73 HardIB · P1 TZ0 November 2018

Consider the triangle $OAB$ , such that $O$ has coordinates $(0, 0, 0)$ , $A$ has coordinates $(0, 1, 2)$ and $B$ has coordinates $(2 b, 0, b - 1)$ , where $b < 0$ .

a) Find, in terms of $b$ , the Cartesian equation of the plane $Π$ that contains this triangle.

Let $M$ be the midpoint of the line segment $[OB]$ .

b) Find, in terms of $b$ , the equation of the line $L$ which passes through $M$ and is perpendicular to the plane $Π$ .

c) Show that $L$ does not intersect the $y$ -axis for any negative value of $b$ .

Show answer

Answer: a) $Π : (b - 1) x + 4 b y - 2 b z = 0$ ; b) $L : x = b + (b - 1) t, y = 4 b t, z = \frac{b - 1}{2} - 2 b t$ . Part c) is proved.

Show working and answer

a) 1. Vectors in the plane: $O A = (0, 1, 2)$ and $O B = (2 b, 0, b - 1)$ . 2. Normal: $O A \times O B = 1 \cdot (b - 1) - 2 \cdot 0 2 \cdot 2 b - 0 \cdot (b - 1) 0 \cdot 0 - 1 \cdot 2 b = b - 1 4 b - 2 b$ . 3. The plane passes through $O$ , so $Π : (b - 1) x + 4 b y - 2 b z = 0$ .

b) 1. $M = (b, 0, \frac{b - 1}{2})$ . 2. $L$ has the direction of the normal $(b - 1, 4 b, - 2 b)$ : $L : ⎩ ⎨ ⎧ x = b + (b - 1) t y = 4 b t z = \frac{b - 1}{2} - 2 b t t \in R .$

c) 1. The $y$ -axis consists of the points with $x = 0$ and $z = 0$ . 2. $x = 0$ gives $t = \frac{- b}{b - 1}$ (we can divide because $b < 0$ implies $b - 1 \neq = 0$ ). 3. Substituting into $z = 0$ : $\frac{b - 1}{2} + \frac{2 b ^{2}}{b - 1} = 0$ . Multiplying by $2 (b - 1)$ : $(b - 1)^{2} + 4 b^{2} = 0$ , that is $5 b^{2} - 2 b + 1 = 0$ . 4. The discriminant is $4 - 20 = - 16 < 0$ , so there are no real solutions. (Another way: $(b - 1)^{2} + 4 b^{2}$ is a sum of squares which would only be zero if $b = 1$ and $b = 0$ at the same time, which is impossible.) 5. There is no value of $b$ for which $L$ intersects the $y$ -axis; in particular, none that is negative.

Answer: a) $Π : (b - 1) x + 4 b y - 2 b z = 0$ ; b) $L : x = b + (b - 1) t, y = 4 b t, z = \frac{b - 1}{2} - 2 b t$ . Part c) is proved.

Exercise 74 HardIB · P1 TZ1 May 2018

The following figure shows a square-based pyramid with vertices $O (0, 0, 0)$ , $A (1, 0, 0)$ , $B (1, 1, 0)$ , $C (0, 1, 0)$ and $D (0, 0, 1)$ .

a) Find the Cartesian equation of the plane $Π_{1}$ , which passes through the points $A$ , $B$ and $D$ .

The Cartesian equation of the plane $Π_{2}$ , which passes through the points $B$ , $C$ and $D$ , is $y + z = 1$ .

b) Find the angle between the faces $ABD$ and $BCD$ .

The plane $Π_{3}$ passes through $O$ and is normal to the line $BD$ .

c) Find the Cartesian equation of $Π_{3}$ .

$Π_{3}$ cuts $AD$ and $BD$ at the points $P$ and $Q$ respectively.

d) Show that $P$ is the midpoint of $AD$ .

e) Find the area of the triangle $OPQ$ .

Show answer

Answer: a) $Π_{1} : x + z = 1$ ; b) $6 0^{\circ}$ ; c) $Π_{3} : - x - y + z = 0$ ; d) $P (\frac{1}{2}, 0, \frac{1}{2})$ is the midpoint of $A D$ , and $Q (\frac{1}{3}, \frac{1}{3}, \frac{2}{3})$ ; e) area $= \frac{3}{12}$

Show working and answer

a) 1. $A B = (0, 1, 0)$ and $A D = (- 1, 0, 1)$ . 2. Normal: $A B \times A D = 1 \cdot 1 - 0 \cdot 0 0 \cdot (- 1) - 0 \cdot 1 0 \cdot 0 - 1 \cdot (- 1) = 101$ . 3. $x + z = d$ with $A (1, 0, 0)$ : $d = 1$ . We check with $B$ and $D$ : $1 + 0 = 1$ and $0 + 1 = 1$ . 4. $Π_{1} : x + z = 1$ .

b) 1. Normals: $n_{1} = (1, 0, 1)$ of $Π_{1}$ and $n_{2} = (0, 1, 1)$ of $Π_{2}$ . 2. $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣} = \frac{1}{2 2} = \frac{1}{2}$ . 3. The angle between the planes is $θ = 6 0^{\circ}$ .

c) 1. $B D = D - B = (- 1, - 1, 1)$ is the normal of $Π_{3}$ . 2. $Π_{3}$ passes through $O$ : $- x - y + z = 0$ .

d) 1. A point on $AD$ is $(1 - s, 0, s)$ . 2. Imposing it on $Π_{3}$ : $- (1 - s) - 0 + s = 0 \Rightarrow 2 s = 1 \Rightarrow s = \frac{1}{2}$ . 3. $P = (\frac{1}{2}, 0, \frac{1}{2})$ , which is the midpoint of $A (1, 0, 0)$ and $D (0, 0, 1)$ .

e) 1. A point on $BD$ is $(1 - t, 1 - t, t)$ . On $Π_{3}$ : $- (1 - t) - (1 - t) + t = 0 \Rightarrow 3 t = 2 \Rightarrow t = \frac{2}{3}$ . 2. $Q = (\frac{1}{3}, \frac{1}{3}, \frac{2}{3})$ . 3. $O P \times O Q = \frac{1}{2} 0 \frac{1}{2} \times \frac{1}{3} \frac{1}{3} \frac{2}{3} = 0 \cdot \frac{2}{3} - \frac{1}{2} \cdot \frac{1}{3} \frac{1}{2} \cdot \frac{1}{3} - \frac{1}{2} \cdot \frac{2}{3} \frac{1}{2} \cdot \frac{1}{3} - 0 = - \frac{1}{6} - \frac{1}{6} \frac{1}{6}$ . 4. Area $= \frac{1}{2} O P \times O Q = \frac{1}{2} \cdot \frac{3}{6} = \frac{3}{12}$ .

Exercise 75 HardIB · P2 TZ2 May 2017

The points A, B and C have the following position vectors relative to the origin O.

$O A = 2 i + j - 2 k$

$O B = 2 i - j + 2 k$

$O C = i + 3 j + 3 k$

a) Find a vector equation of the line (BC).

b) Determine whether the lines (OA) and (BC) intersect or not.

c) Find the Cartesian equation of the plane $Π_{1}$ , which passes through C and is perpendicular to $O A$ .

d) Show that the line (BC) lies in the plane $Π_{1}$ .

The plane $Π_{2}$ contains the points O, A and B, and the plane $Π_{3}$ contains the points O, A and C.

e) Verify that $2 j + k$ is perpendicular to the plane $Π_{2}$ .

f) Find a vector which is perpendicular to the plane $Π_{3}$ .

g) Find the acute angle between the planes $Π_{2}$ and $Π_{3}$ .

Show answer

Answer: a) $r = 2 - 1 2 + λ - 1 41$ ; b) they do not intersect; c) $2 x + y - 2 z + 1 = 0$ ; f) $9 i - 8 j + 5 k$ ; g) $α \approx 67. 8^{\circ}$ . Parts d) and e) are proofs.

Show working and answer

a) 1. $B C = C - B = (1 - 2, 3 + 1, 3 - 2) = (- 1, 4, 1)$ . 2. Taking $B$ as the point: $r = 2 - 1 2 + λ - 1 41$ , $λ \in R$ .

b) 1. The line $(O A)$ is $r = s 21 - 2$ . Its direction vectors $(2, 1, - 2)$ and $(- 1, 4, 1)$ are not proportional, so they are not parallel. 2. We look for a common point: $2 s = 2 - λ$ , $s = - 1 + 4 λ$ , $- 2 s = 2 + λ$ . 3. Substituting $s = - 1 + 4 λ$ into the first: $- 2 + 8 λ = 2 - λ$ , so $λ = \frac{4}{9}$ and $s = \frac{7}{9}$ . 4. We check in the third equation: $- 2 s = - \frac{14}{9}$ , but $2 + λ = \frac{22}{9}$ . They do not match. 5. The system has no solution: the lines do not intersect (they are skew).

c) 1. The normal vector is $O A = (2, 1, - 2)$ , so $Π_{1} : 2 x + y - 2 z = d$ . 2. Through $C (1, 3, 3)$ : $d = 2 + 3 - 6 = - 1$ . 3. $Π_{1} : 2 x + y - 2 z + 1 = 0$ .

d) 1. $B (2, - 1, 2)$ lies in $Π_{1}$ : $4 - 1 - 4 + 1 = 0$ . And $C (1, 3, 3)$ also lies in it: $2 + 3 - 6 + 1 = 0$ . 2. As two points of the line $(B C)$ lie in the plane, the whole line is contained in it. (Another way: $B \in Π_{1}$ and $B C \cdot O A = - 2 + 4 - 2 = 0$ .)

e) 1. A normal of $Π_{2}$ is $O A \times O B$ : $21 - 2 \times 2 - 1 2 = (1) (2) - (- 2) (- 1) (- 2) (2) - (2) (2) (2) (- 1) - (1) (2) = 0 - 8 - 4 = - 4021 .$ 2. It is proportional to $2 j + k$ , which is therefore perpendicular to $Π_{2}$ . (Another way: $(2 j + k) \cdot O A = 2 - 2 = 0$ and $(2 j + k) \cdot O B = - 2 + 2 = 0$ .)

f) 1. $O A \times O C = 21 - 2 \times 133 = (1) (3) - (- 2) (3) (- 2) (1) - (2) (3) (2) (3) - (1) (1) = 9 - 8 5$ . 2. A vector perpendicular to $Π_{3}$ is $9 i - 8 j + 5 k$ .

g) 1. Normals: $n_{2} = (0, 2, 1)$ and $n_{3} = (9, - 8, 5)$ . 2. $n_{2} \cdot n_{3} = - 16 + 5 = - 11$ , $∣ n_{2} ∣ = 5$ , $∣ n_{3} ∣ = 170$ . 3. $cos α = \frac{11}{5 170} = \frac{11}{850}$ (absolute value, for the acute angle), so $α \approx 67. 8^{\circ}$ .

Answer: a) $r = 2 - 1 2 + λ - 1 41$ ; b) they do not intersect; c) $2 x + y - 2 z + 1 = 0$ ; f) $9 i - 8 j + 5 k$ ; g) $α \approx 67. 8^{\circ}$ . Parts d) and e) are proofs.

Exercise 76 HardIB · P2 TZ2 May 2019

The plane $Π_{1}$ contains the points $P (1, 6, - 7)$ , $Q (0, 1, 1)$ and $R (2, 0, - 4)$ .

a) Find the Cartesian equation of the plane that contains P, Q and R.

The Cartesian equation of the plane $Π_{2}$ is $x - 3 y - z = 3$ .

b) Given that $Π_{1}$ and $Π_{2}$ intersect in a line $L$ , verify that the vector equation of $L$ can be given by $r = \frac{5}{4} 0 - \frac{7}{4} + λ \frac{1}{2} 1 - \frac{5}{2}$ .

The Cartesian equation of the plane $Π_{3}$ is $a x + b y + cz = 1$ .

c) Given that $Π_{3}$ is parallel to the line $L$ , show that $a + 2 b - 5 c = 0$ .

Now consider the case in which $Π_{3}$ contains $L$ .

d) (i) Show that $5 a - 7 c = 4$ .

(ii) Given that $Π_{3}$ is equally inclined to $Π_{1}$ as to $Π_{2}$ , determine two possible and distinct Cartesian equations for $Π_{3}$ .

Show answer

Answer: a) $3 x + y + z = 2$ ; d) (ii) $Π_{3} : 4 x - 2 y = 5$ and $Π_{3} : - 2 x - 4 y - 2 z = 1$ . Parts b), c) and d) (i) are proofs.

Show working and answer

a) 1. $P Q = (- 1, - 5, 8)$ and $P R = (1, - 6, 3)$ . 2. Normal: $P Q \times P R = (- 5) (3) - (8) (- 6) (8) (1) - (- 1) (3) (- 1) (- 6) - (- 5) (1) = 331111 = 11311$ . 3. The plane is $3 x + y + z = d$ . Through $P$ : $d = 3 + 6 - 7 = 2$ . We check with $Q$ : $0 + 1 + 1 = 2$ and with $R$ : $6 + 0 - 4 = 2$ . 4. $Π_{1} : 3 x + y + z = 2$ .

b) 1. The point $(\frac{5}{4}, 0, - \frac{7}{4})$ lies on $Π_{1}$ : $\frac{15}{4} + 0 - \frac{7}{4} = 2$ ; and on $Π_{2}$ : $\frac{5}{4} - 0 + \frac{7}{4} = 3$ . 2. The vector $(\frac{1}{2}, 1, - \frac{5}{2})$ is parallel to both planes: $3 \cdot \frac{1}{2} + 1 - \frac{5}{2} = 0$ and $\frac{1}{2} - 3 + \frac{5}{2} = 0$ . 3. A common point and a direction parallel to both planes determine the line of intersection: the given equation is correct.

c) 1. $Π_{3}$ is parallel to $L$ if its normal $(a, b, c)$ is perpendicular to the direction of $L$ . 2. $(a, b, c) \cdot (\frac{1}{2}, 1, - \frac{5}{2}) = 0$ , i.e. $\frac{a}{2} + b - \frac{5 c}{2} = 0$ . Multiplying by 2: $a + 2 b - 5 c = 0$ .

d) (i) 1. If $Π_{3}$ contains $L$ , the point $(\frac{5}{4}, 0, - \frac{7}{4})$ of $L$ lies on $Π_{3}$ : $\frac{5 a}{4} - \frac{7 c}{4} = 1$ . 2. Multiplying by 4: $5 a - 7 c = 4$ .

(ii) 1. Being equally inclined to $Π_{1}$ as to $Π_{2}$ means forming the same angle with both: $\frac{∣ n \cdot n _{1} ∣}{∣ n ∣ ∣ n _{1} ∣} = \frac{∣ n \cdot n _{2} ∣}{∣ n ∣ ∣ n _{2} ∣}$ , with $n = (a, b, c)$ , $n_{1} = (3, 1, 1)$ and $n_{2} = (1, - 3, - 1)$ . 2. As $∣ n_{1} ∣ = ∣ n_{2} ∣ = 11$ , this gives $∣ 3 a + b + c ∣ = ∣ a - 3 b - c ∣$ . 3. First case, $3 a + b + c = a - 3 b - c$ : $a + 2 b + c = 0$ . Together with $a + 2 b - 5 c = 0$ (part c) we obtain $6 c = 0$ , $c = 0$ and $a = - 2 b$ . Then $5 a = 4$ gives $a = \frac{4}{5}$ , $b = - \frac{2}{5}$ . 4. Second case, $3 a + b + c = - (a - 3 b - c)$ : $4 a - 2 b = 0$ , i.e. $b = 2 a$ . With $a + 2 b - 5 c = 0$ we obtain $5 a - 5 c = 0$ , $c = a$ . Then $5 a - 7 a = 4$ gives $a = - 2$ , $b = - 4$ , $c = - 2$ . 5. The two planes are $\frac{4}{5} x - \frac{2}{5} y = 1$ , i.e. $4 x - 2 y = 5$ , and $- 2 x - 4 y - 2 z = 1$ .

Answer: a) $3 x + y + z = 2$ ; d) (ii) $Π_{3} : 4 x - 2 y = 5$ and $Π_{3} : - 2 x - 4 y - 2 z = 1$ . Parts b), c) and d) (i) are proofs.

Exercise 77 HardIB · P2 TZ1 May 2004

a) The point $P (1, 2, 11)$ lies on the plane $π_{1}$ . The vector $3 i - 4 j + k$ is perpendicular to $π_{1}$ . Find the Cartesian equation of $π_{1}$ .

b) The plane $π_{2}$ has equation $x + 3 y - z = - 4$ .

(i) Prove that the point $P$ also lies on the plane $π_{2}$ .

(ii) Find a vector equation of the line of intersection of $π_{1}$ and $π_{2}$ .

c) Calculate the acute angle between $π_{1}$ and $π_{2}$ .

Show answer

Answer: (a) $π_{1} : 3 x - 4 y + z - 6 = 0$ ; (b)(ii) $r = 1211 + t 1413$ ; (c) $θ \approx 53. 7^{\circ}$ . Part (b)(i) is verified.

Show working and answer

a) 1. The normal is $(3, - 4, 1)$ , so $π_{1} : 3 x - 4 y + z = d$ . 2. As $P (1, 2, 11) \in π_{1}$ : $d = 3 - 8 + 11 = 6$ . 3. $π_{1} : 3 x - 4 y + z = 6$ .

b) (i) 1. Substituting $P$ into $π_{2}$ : $1 + 3 \cdot 2 - 11 = - 4$ . It holds, so $P \in π_{2}$ .

(ii) 1. The line of intersection passes through $P$ (it lies in both planes) and is perpendicular to both normals, $(3, - 4, 1)$ and $(1, 3, - 1)$ . Its direction is $3 - 4 1 \times 13 - 1 = (- 4) (- 1) - 1 \cdot 3 1 \cdot 1 - 3 (- 1) 3 \cdot 3 - (- 4) \cdot 1 = 1413 .$ 2. A vector equation is $r = 1211 + t 1413 .$

c) 1. The acute angle between the planes is the acute angle between their normals: $cos θ = \frac{∣3 \cdot 1 + ( - 4 ) \cdot 3 + 1 \cdot ( - 1 ) ∣}{26 11} = \frac{10}{286} \approx 0.5913.$ 2. $θ = arccos \frac{10}{286} \approx 53. 7^{\circ}$ .

Answer: (a) $π_{1} : 3 x - 4 y + z - 6 = 0$ ; (b)(ii) $r = 1211 + t 1413$ ; (c) $θ \approx 53. 7^{\circ}$ . Part (b)(i) is verified.

Exercise 78 HardIB · P2 TZ1 May 2005

A plane $π_{1}$ has equation $r \cdot 654 = 15$ .

a) A point $P (p, - p, p)$ lies on the plane $π_{1}$ , and $Q$ is the point where the plane $π_{1}$ cuts the $y$ -axis.

(i) Calculate the coordinates of $P$ and of $Q$ .

(ii) Show that $P Q$ is parallel to the vector $u$ , where $u = i - 2 j + k$ .

b) Another plane $π_{2}$ cuts $π_{1}$ in the line $(P Q)$ . The point $T (1, 0, - 1)$ lies on $π_{2}$ .

(i) Find the equation of $π_{2}$ , giving your answer in the form $A x + B y + C z + D = 0$ .

(ii) Find the angle between $π_{1}$ and $π_{2}$ .

Show answer

Answer: (a)(i) $P (3, - 3, 3)$ , $Q (0, 3, 0)$ ; (b)(i) $π_{2} : 5 x + 2 y - z - 6 = 0$ ; (b)(ii) $θ \approx 41. 5^{\circ}$ . Part (a)(ii) is verified.

Show working and answer

The equation of $π_{1}$ is $6 x + 5 y + 4 z = 15$ .

a) (i) 1. For $P (p, - p, p)$ : $6 p - 5 p + 4 p = 15 \Rightarrow 5 p = 15 \Rightarrow p = 3$ . Thus, $P (3, - 3, 3)$ . 2. The $y$ -axis consists of the points $(0, y, 0)$ . In $π_{1}$ : $5 y = 15 \Rightarrow y = 3$ . Thus, $Q (0, 3, 0)$ .

(ii) 1. $P Q = Q - P = - 3 6 - 3 = - 3 1 - 2 1 = - 3 u$ . 2. It is a multiple of $u$ , so $P Q ∥ u$ .

b) (i) 1. Plane $π_{2}$ contains the line $P Q$ (direction $u$ ) and the point $T$ . With $Q (0, 3, 0)$ and $T (1, 0, - 1)$ , a second vector in the plane is $QT = (1, - 3, - 1)$ . 2. The normal of $π_{2}$ is $u \times QT = 1 - 2 1 \times 1 - 3 - 1 = (- 2) (- 1) - 1 \cdot (- 3) 1 \cdot 1 - 1 \cdot (- 1) 1 \cdot (- 3) - (- 2) \cdot 1 = 52 - 1 .$ 3. Then $π_{2} : 5 x + 2 y - z + D = 0$ . As $T (1, 0, - 1) \in π_{2}$ : $5 + 0 + 1 + D = 0 \Rightarrow D = - 6$ . 4. $π_{2} : 5 x + 2 y - z - 6 = 0$ . Check with $P$ : $15 - 6 - 3 - 6 = 0$ , and with $Q$ : $0 + 6 - 0 - 6 = 0$ .

(ii) 1. The angle between the planes is the one formed by their normals $(6, 5, 4)$ and $(5, 2, - 1)$ : $cos θ = \frac{∣30 + 10 - 4∣}{77 30} = \frac{36}{2310} \approx 0.7490.$ 2. $θ = arccos \frac{36}{2310} \approx 41. 5^{\circ}$ .

Answer: (a)(i) $P (3, - 3, 3)$ , $Q (0, 3, 0)$ ; (b)(i) $π_{2} : 5 x + 2 y - z - 6 = 0$ ; (b)(ii) $θ \approx 41. 5^{\circ}$ . Part (a)(ii) is verified.

Exercise 79 HardIB · P2 TZ1 May 2007

Two planes $π_{1}$ and $π_{2}$ are represented by the equations

$π_{1} : r = 315 + λ - 2 23 + μ 210 π_{2} : 2 x - y - 2 z = 4.$

a) (i) Calculate $- 2 23 \times 210$ .

(ii) Show that the equation of $π_{1}$ can be written as $x - 2 y + 2 z = 11$ .

b) Show that the plane $π_{1}$ is perpendicular to $π_{2}$ .

c) The line $l_{1}$ is the line of intersection of $π_{1}$ and $π_{2}$ . Find the equation of $l_{1}$ , giving your answer in parametric form.

d) The line $l_{2}$ is parallel to both $π_{1}$ and $π_{2}$ , and passes through $P (3, - 5, - 1)$ . Find an equation for $l_{2}$ in Cartesian form.

e) Let $Q$ be the foot of the perpendicular from $P$ to the plane $π_{2}$ .

(i) Find the coordinates of $Q$ .

(ii) Calculate $P Q$ .

Show answer

Answer: a) (i) $- 3 i + 6 j - 6 k$ ; (ii) $x - 2 y + 2 z = 11$ . b) $n_{1} \cdot n_{2} = 0$ . c) $x = t, y = - 5 + t, z = \frac{1}{2} + \frac{1}{2} t$ . d) $\frac{x - 3}{2} = \frac{y + 5}{2} = \frac{z + 1}{1}$ . e) (i) $Q (1, - 4, 1)$ ; (ii) $P Q = 3$ .

Show working and answer

a) (i) 1. We apply the formula for the vector (cross) product: $- 2 23 \times 210 = 2 \cdot 0 - 3 \cdot 1 3 \cdot 2 - (- 2) \cdot 0 (- 2) \cdot 1 - 2 \cdot 2 = - 3 6 - 6 = - 3 i + 6 j - 6 k .$

(ii) 1. This vector is perpendicular to both direction vectors of the plane, so it is normal to $π_{1}$ . A simpler normal vector is $1 - 2 2$ (dividing by $- 3$ ). 2. Hence $π_{1} : x - 2 y + 2 z = d$ . Since it passes through $(3, 1, 5)$ : $d = 3 - 2 + 10 = 11$ . 3. Therefore $π_{1} : x - 2 y + 2 z = 11$ .

b) 1. The normals are $n_{1} = 1 - 2 2$ and $n_{2} = 2 - 1 - 2$ . 2. $n_{1} \cdot n_{2} = 2 + 2 - 4 = 0$ , so the planes are perpendicular.

c) 1. The points of $l_{1}$ satisfy both equations simultaneously: $x - 2 y + 2 z = 11, 2 x - y - 2 z = 4.$ 2. Adding them eliminates $z$ : $3 x - 3 y = 15$ , that is, $y = x - 5$ . 3. We take $x = t$ as the parameter: $y = - 5 + t$ . Substituting into the first: $t - 2 (- 5 + t) + 2 z = 11$ , so $2 z = 1 + t$ and $z = \frac{1}{2} + \frac{1}{2} t$ . 4. Parametric equations of $l_{1}$ : $x = t, y = - 5 + t, z = \frac{1}{2} + \frac{1}{2} t .$

d) 1. A line parallel to two planes has as its direction a vector perpendicular to both normals: $n_{1} \times n_{2} = (- 2) (- 2) - 2 \cdot (- 1) 2 \cdot 2 - 1 \cdot (- 2) 1 \cdot (- 1) - (- 2) \cdot 2 = 663 ∥ 221 .$ 2. It passes through $P (3, - 5, - 1)$ , so its Cartesian equation is $\frac{x - 3}{2} = \frac{y + 5}{2} = \frac{z + 1}{1} .$

e) (i) 1. The line through $P$ perpendicular to $π_{2}$ has direction $n_{2}$ : $r = 3 - 5 - 1 + s 2 - 1 - 2$ . 2. We impose that the point lies on $π_{2}$ : $2 (3 + 2 s) - (- 5 - s) - 2 (- 1 - 2 s) = 4$ , that is, $6 + 4 s + 5 + s + 2 + 4 s = 4$ , i.e. $13 + 9 s = 4$ and $s = - 1$ . 3. $Q = (3 - 2, - 5 + 1, - 1 + 2) = (1, - 4, 1)$ .

(ii) 1. $P Q = s n_{2}$ with $s = - 1$ , so $P Q = ∣ - 1 ∣ \cdot ∣ n_{2} ∣ = 4 + 1 + 4 = 3$ .

Exercise 80 HardIB · P2 TZ2 May 2007

Consider the vectors $a = i - j + k$ , $b = i + 2 j + 4 k$ and $c = 2 i - 5 j - k$ .

a) Given that $c = m a + n b$ , where $m, n \in Z$ , find the values of $m$ and $n$ .

b) Find a unit vector $u$ that is normal to $a$ and also to $b$ .

c) The plane $π_{1}$ contains the point $A (1, - 1, 1)$ and is normal to the vector $b$ .

The plane cuts the $x$ , $y$ and $z$ axes at the points $L$ , $M$ and $N$ respectively.

(i) Find the Cartesian equation of $π_{1}$ .

(ii) Write down the coordinates of $L$ , $M$ and $N$ .

d) The normal line to $π_{1}$ through the origin meets $π_{1}$ at the point $P$ .

(i) Find the coordinates of $P$ .

(ii) Hence find the distance from $π_{1}$ to the origin.

e) The plane $π_{2}$ has equation $x + 2 y + 4 z = 4$ .

Calculate the angle between $π_{2}$ and a line parallel to $a$ .

Show answer

Answer: a) $m = 3, n = - 1$ . b) $u = \frac{1}{54} - 6 - 3 3$ . c) (i) $x + 2 y + 4 z - 3 = 0$ ; (ii) $L (3, 0, 0)$ , $M (0, \frac{3}{2}, 0)$ , $N (0, 0, \frac{3}{4})$ . d) (i) $P (\frac{1}{7}, \frac{2}{7}, \frac{4}{7})$ ; (ii) $\frac{21}{7}$ . e) $θ \approx 22. 2^{\circ}$ .

Show working and answer

a) 1. We equate components of $c = m a + n b$ : $m + n = 2, - m + 2 n = - 5, m + 4 n = - 1.$ 2. Adding the first two: $3 n = - 3$ , so $n = - 1$ and $m = 3$ . 3. We check with the third: $3 + 4 (- 1) = - 1$ . It holds.

b) 1. A vector perpendicular to $a$ and $b$ is their vector (cross) product: $a \times b = (- 1) \cdot 4 - 1 \cdot 2 1 \cdot 1 - 1 \cdot 4 1 \cdot 2 - (- 1) \cdot 1 = - 6 - 3 3 .$ 2. Its magnitude is $36 + 9 + 9 = 54 = 36$ . 3. We divide by the magnitude: $u = \frac{1}{54} - 6 - 3 3 = \frac{1}{6} - 2 - 1 1 .$ (The opposite vector is also valid.)

c) (i) 1. Since $π_{1}$ is normal to $b$ , its equation is $x + 2 y + 4 z = d$ . 2. It passes through $A (1, - 1, 1)$ : $d = 1 - 2 + 4 = 3$ . 3. $π_{1} : x + 2 y + 4 z - 3 = 0$ .

(ii) 1. Intersection with the $x$ -axis ( $y = z = 0$ ): $x = 3$ , so $L (3, 0, 0)$ . 2. Intersection with the $y$ -axis ( $x = z = 0$ ): $2 y = 3$ , so $M (0, \frac{3}{2}, 0)$ . 3. Intersection with the $z$ -axis ( $x = y = 0$ ): $4 z = 3$ , so $N (0, 0, \frac{3}{4})$ .

d) (i) 1. The normal line through the origin is $(x, y, z) = (t, 2 t, 4 t)$ . 2. We substitute into $π_{1}$ : $t + 4 t + 16 t = 3$ , that is, $21 t = 3$ and $t = \frac{1}{7}$ . 3. $P = (\frac{1}{7}, \frac{2}{7}, \frac{4}{7})$ .

(ii) The distance from the origin to $π_{1}$ is the length of the perpendicular segment $O P$ : $∣ O P ∣ = \frac{1}{49} + \frac{4}{49} + \frac{16}{49} = \frac{21}{7} .$

e) 1. The angle $θ$ between a line with direction $a$ and the plane with normal $n = 124$ satisfies $sin θ = \frac{∣ a \cdot n ∣}{∣ a ∣ ∣ n ∣}$ . 2. $a \cdot n = 1 - 2 + 4 = 3$ , $∣ a ∣ = 3$ and $∣ n ∣ = 21$ . 3. $sin θ = \frac{3}{3 21} = \frac{3}{63} = \frac{1}{7}$ , so $θ = arcsin \frac{1}{7} \approx 22. 2^{\circ}$ .

Exercise 81 HardIB · P2 TZ1 May 2010

A plane $π$ has a vector equation

$r = (- 2 i + 3 j - 2 k) + λ (2 i + 3 j + 2 k) + μ (6 i - 3 j + 2 k) .$

a) Show that the Cartesian equation of the plane $π$ is $3 x + 2 y - 6 z = 12$ .

b) The plane $π$ cuts the $x$ , $y$ and $z$ axes at $A$ , $B$ and $C$ , respectively.

Find the coordinates of $A$ , $B$ and $C$ .

c) Find the volume of the pyramid $O A B C$ .

d) Find the angle between the plane $π$ and the $x$ -axis.

e) Using the above, or otherwise, calculate the distance from the origin to the plane $π$ .

f) Using your answers to (c) and (e), find the area of triangle $A B C$ .

Show answer

Answer: a) $3 x + 2 y - 6 z = 12$ . b) $A (4, 0, 0)$ , $B (0, 6, 0)$ , $C (0, 0, - 2)$ . c) $V = 8$ . d) $θ \approx 25. 4^{\circ}$ . e) $\frac{12}{7}$ . f) $14$ .

Show working and answer

a) 1. A normal vector is the vector (cross) product of the two direction vectors: $232 \times 6 - 3 2 = 3 \cdot 2 - 2 \cdot (- 3) 2 \cdot 6 - 2 \cdot 2 2 \cdot (- 3) - 3 \cdot 6 = 128 - 24 = 4 32 - 6 .$ 2. Hence $π : 3 x + 2 y - 6 z = d$ . The point $(- 2, 3, - 2)$ lies on the plane: $d = - 6 + 6 + 12 = 12$ . 3. Therefore $π : 3 x + 2 y - 6 z = 12$ .

b) 1. $x$ -axis ( $y = z = 0$ ): $3 x = 12$ , $x = 4$ , so $A (4, 0, 0)$ . 2. $y$ -axis ( $x = z = 0$ ): $2 y = 12$ , $y = 6$ , so $B (0, 6, 0)$ . 3. $z$ -axis ( $x = y = 0$ ): $- 6 z = 12$ , $z = - 2$ , so $C (0, 0, - 2)$ .

c) 1. $O A$ , $O B$ and $O C$ are mutually perpendicular, so the volume of the pyramid is $\frac{1}{3} \cdot area of the base \cdot height$ , with base $O A B$ and height $∣ O C ∣$ . 2. Area of $O A B$ : $\frac{1}{2} \cdot 4 \cdot 6 = 12$ . Height: $∣ O C ∣ = 2$ . 3. $V = \frac{1}{3} \cdot 12 \cdot 2 = 8$ .

d) 1. The angle $θ$ between the plane and the $x$ -axis (direction $i$ ) satisfies $sin θ = \frac{∣ i \cdot n ∣}{∣ i ∣ ∣ n ∣}$ with $n = 32 - 6$ . 2. $∣ n ∣ = 9 + 4 + 36 = 7$ and $i \cdot n = 3$ , so $sin θ = \frac{3}{7}$ . 3. $θ = arcsin \frac{3}{7} \approx 25. 4^{\circ}$ .

e) 1. Distance from the origin to the plane $3 x + 2 y - 6 z - 12 = 0$ : $d = \frac{∣3 \cdot 0 + 2 \cdot 0 - 6 \cdot 0 - 12∣}{3 ^{2} + 2 ^{2} + 6 ^{2}} = \frac{12}{7} .$

f) 1. We take the triangle $A B C$ as base and the distance $d = \frac{12}{7}$ from the origin to the plane as height: $V = \frac{1}{3} \cdot Area (A B C) \cdot d$ . 2. With $V = 8$ : $8 = \frac{1}{3} \cdot Area (A B C) \cdot \frac{12}{7} = \frac{4}{7} Area (A B C)$ . 3. $Area (A B C) = 8 \cdot \frac{7}{4} = 14$ .

Answer: a) $3 x + 2 y - 6 z = 12$ . b) $A (4, 0, 0)$ , $B (0, 6, 0)$ , $C (0, 0, - 2)$ . c) $V = 8$ . d) $θ \approx 25. 4^{\circ}$ . e) $\frac{12}{7}$ . f) $14$ .

Exercise 82 HardIB · P2 TZ0 November 2021

Three points $A (3, 0, 0)$ , $B (0, - 2, 0)$ and $C (1, 1, - 7)$ lie on the plane $Π_{1}$ .

a) i) Calculate the vector $A B$ and the vector $A C$ .

(ii) Hence, find the equation of $Π_{1}$ , giving your answer in the form $a x + b y + cz = d$ , where $a, b, c, d \in Z$ .

The plane $Π_{2}$ has equation $3 x - y + 2 z = 2$ .

b) The line $L$ is the intersection of $Π_{1}$ and $Π_{2}$ . Verify that the vector equation of $L$ can be written in the form $r = 0 - 2 0 + λ 11 - 1$ .

c) Let $Π_{3}$ be the plane $2 x - 2 z = 3$ .

The line $L$ and the plane $Π_{3}$ intersect at the point $P$ .

(i) Show that at $P$ , $λ = \frac{3}{4}$ .

(ii) Hence, calculate the coordinates of $P$ .

d) The point $B (0, - 2, 0)$ lies on $L$ .

(i) Find the reflection of the point $B$ in the plane $Π_{3}$ .

(ii) Hence, find the vector equation of the line formed when $L$ is reflected in $Π_{3}$ .

Show answer

Answer: a) i) $A B = - 3 - 2 0$ , $A C = - 2 1 - 7$ ; ii) $2 x - 3 y - z = 6$ ; c) ii) $P (\frac{3}{4}, - \frac{5}{4}, - \frac{3}{4})$ ; d) i) $B^{'} (\frac{3}{2}, - 2, - \frac{3}{2})$ ; ii) $r = \frac{3}{2} - 2 - \frac{3}{2} + t 1 - 1 - 1$ .

Show working and answer

a) i) $A B = B - A = - 3 - 2 0$ and $A C = C - A = - 2 1 - 7$ .

(ii) 1. Normal to the plane: $A B \times A C = (- 2) (- 7) - 0 \cdot 1 0 \cdot (- 2) - (- 3) (- 7) (- 3) (1) - (- 2) (- 2) = 14 - 21 - 7 = 7 2 - 3 - 1 .$ 2. The plane is $2 x - 3 y - z = d$ ; with $A$ : $d = 6$ . Check with $B$ : $6$ , and with $C$ : $2 - 3 + 7 = 6$ . 3. $Π_{1} : 2 x - 3 y - z = 6$ .

b) 1. The point $B (0, - 2, 0)$ lies on $Π_{1}$ ( $6 = 6$ ) and on $Π_{2}$ ( $2 = 2$ ). 2. A general point on the given line is $(λ, - 2 + λ, - λ)$ . On $Π_{1}$ : $2 λ - 3 (- 2 + λ) - (- λ) = 2 λ + 6 - 3 λ + λ = 6$ . This holds for all $λ$ . 3. On $Π_{2}$ : $3 λ - (- 2 + λ) + 2 (- λ) = 3 λ + 2 - λ - 2 λ = 2$ . This also holds for all $λ$ . 4. All points of that line lie on both planes and its direction is not zero; since the intersection of two non-parallel planes is a line, this line is $L$ .

c) i) 1. Substitute the general point $(λ, - 2 + λ, - λ)$ into $Π_{3}$ : $2 λ - 2 (- λ) = 3$ . 2. This gives $4 λ = 3$ , i.e. $λ = \frac{3}{4}$ .

(ii) With $λ = \frac{3}{4}$ : $P = (\frac{3}{4}, - 2 + \frac{3}{4}, - \frac{3}{4}) = (\frac{3}{4}, - \frac{5}{4}, - \frac{3}{4})$ .

d) i) 1. The normal of $Π_{3}$ is $(2, 0, - 2)$ , parallel to $(1, 0, - 1)$ . The line perpendicular to $Π_{3}$ through $B$ is $(s, - 2, - s)$ . 2. Intersect with $Π_{3}$ : $2 s - 2 (- s) = 3$ , so $s = \frac{3}{4}$ . The foot of the perpendicular is $M = (\frac{3}{4}, - 2, - \frac{3}{4})$ . 3. The reflection $B^{'}$ is the point symmetric to $B$ with respect to $M$ : $B^{'} = 2 M - B = (\frac{3}{2}, - 2, - \frac{3}{2})$ .

(ii) 1. $P$ lies on $Π_{3}$ , so it does not change under the reflection. The reflected line passes through $P$ and $B^{'}$ . 2. $P B^{'} = \frac{3}{2} - \frac{3}{4} - 2 + \frac{5}{4} - \frac{3}{2} + \frac{3}{4} = \frac{3}{4} - \frac{3}{4} - \frac{3}{4} = \frac{3}{4} 1 - 1 - 1$ . 3. Reflected line: $r = \frac{3}{2} - 2 - \frac{3}{2} + t 1 - 1 - 1 .$

Angles between lines and planes

Exercise 83 BasicIB · P1 November 2003

Find the angle between the plane $3 x - 2 y + 4 z = 12$ and the $z$ -axis. Give your answer to the nearest degree.

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Answer: $4 8^{\circ}$

Show working and answer

The normal vector of the plane is $n = (3, - 2, 4)$ , with $∣ n ∣ = 9 + 4 + 16 = 29$ . The $z$ -axis has direction vector $k = (0, 0, 1)$ .
The angle $φ$ between a line and a plane satisfies $sin φ = \frac{∣ n \cdot k ∣}{∣ n ∣∣ k ∣} = \frac{4}{29}$ .
$φ = arcsin \frac{4}{29} \approx 47.9 7^{\circ} \approx 4 8^{\circ}$ .

Answer: $4 8^{\circ}$

Exercise 84 BasicIB · P2 TZ0 November 2014

Consider the two planes

$π_{1} : 4 x + 2 y - z = 8$

$π_{2} : x + 3 y + 3 z = 3$

Find the angle between $π_{1}$ and $π_{2}$ , correct to the nearest whole number of degrees.

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Answer: $6 9^{\circ}$

Show working and answer

The normal vectors are $n_{1} = (4, 2, - 1)$ and $n_{2} = (1, 3, 3)$ .
$n_{1} \cdot n_{2} = 4 + 6 - 3 = 7$ , $∣ n_{1} ∣ = 21$ , $∣ n_{2} ∣ = 19$ .
$cos θ = \frac{7}{21 19} = \frac{7}{399} \approx 0.350$ , so $θ \approx 69. 5^{\circ}$ with the calculator unrounded; to the nearest whole number, $6 9^{\circ}$ .

Answer: $6 9^{\circ}$

Exercise 85 BasicIB · P2 TZ0 November 2016

Find the acute angle between the planes whose equations are $x + y + z = 3$ and $2 x - z = 2$ .

Show answer

Answer: $θ \approx 75. 0^{\circ}$

Show working and answer

The angle between two planes is the angle between their normal vectors. The normals are $n_{1} = (1, 1, 1)$ and $n_{2} = (2, 0, - 1)$ .
$n_{1} \cdot n_{2} = 2 + 0 - 1 = 1$ , $∣ n_{1} ∣ = 3$ and $∣ n_{2} ∣ = 5$ .
$cos θ = \frac{∣ 1 ∣}{3 5} = \frac{1}{15}$ , so $θ = arccos \frac{1}{15} \approx 75. 0^{\circ}$ .

Answer: $θ \approx 75. 0^{\circ}$

Exercise 86 BasicIB · P1 May 1998

Calculate the acute angle between the planes $2 x + 3 y - z = 7$ and $7 x - y + 3 z = - 5$ , giving the angle correct to one decimal place.

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Answer: $θ \approx 73. 8^{\circ}$

Show working and answer

Normal vectors: $n_{1} = (2, 3, - 1)$ and $n_{2} = (7, - 1, 3)$ .
$n_{1} \cdot n_{2} = 14 - 3 - 3 = 8$ , $∣ n_{1} ∣ = 14$ and $∣ n_{2} ∣ = 59$ .
$cos θ = \frac{8}{14 59} = \frac{8}{826}$ , and therefore $θ = arccos \frac{8}{826} \approx 73. 8^{\circ}$ .

Answer: $θ \approx 73. 8^{\circ}$

Exercise 87 BasicIB · P1 TZ0 November 2006

The lines $L_{1}$ and $L_{2}$ are defined by the parametric equations

$L_{1} : x = 1 + 2 λ, y = 1 + 3 λ, z = 1 - λ$

$L_{2} : x = 2 - μ, y = 3 + 4 μ, z = 4 + 2 μ$

Find the angle between $L_{1}$ and $L_{2}$ .

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Answer: $θ \approx 62. 2^{\circ}$

Show working and answer

The direction vectors are $d_{1} = (2, 3, - 1)$ and $d_{2} = (- 1, 4, 2)$ .
Their scalar product is $2 (- 1) + 3 \cdot 4 + (- 1) \cdot 2 = - 2 + 12 - 2 = 8$ , and the magnitudes are $∣ d_{1} ∣ = 14$ and $∣ d_{2} ∣ = 21$ .
Then $cos θ = \frac{8}{14 21} = \frac{8}{7 6} \approx 0.4666 \Rightarrow θ = arccos \frac{8}{7 6} \approx 62. 2^{\circ} .$ Since the cosine is positive, the angle is already acute, and this is the angle between the lines.

Answer: $θ \approx 62. 2^{\circ}$

Exercise 88 BasicIB · P1 TZ2 May 2007

Find the cosine of the angle $θ$ between the planes $π_{1}$ and $π_{2}$ , where $π_{1}$ has equation $- 2 x + y - z = 2$ and $π_{2}$ has equation $x + 2 y - z = 6$ .

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Answer: $cos θ = \frac{1}{6}$ ( $θ \approx 80. 4^{\circ}$ )

Show working and answer

The normals are $n_{1} = (- 2, 1, - 1)$ and $n_{2} = (1, 2, - 1)$ .
$n_{1} \cdot n_{2} = - 2 + 2 + 1 = 1$ , $∣ n_{1} ∣ = 6$ and $∣ n_{2} ∣ = 6$ .
$cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣} = \frac{1}{6}$ , so $θ = arccos \frac{1}{6} \approx 80. 4^{\circ}$ .

Answer: $cos θ = \frac{1}{6}$ ( $θ \approx 80. 4^{\circ}$ )

Exercise 89 BasicIB · P2 TZ1 May 2008

A ray of light coming from the point $(- 1, 3, 2)$ travels in the direction of the vector $41 - 2$ and cuts the plane $π : x + 3 y + 2 z - 24 = 0$ . Calculate the angle that the ray of light makes with the plane.

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Answer: the ray makes an angle of $θ \approx 10. 1^{\circ}$ with the plane.

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The direction of the ray is $d = 41 - 2$ and the normal of the plane is $n = 132$ .
The angle $θ$ between a line and a plane satisfies $sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣ ∣ n ∣}$ (it is the complement of the angle between $d$ and $n$ ).
We calculate: $d \cdot n = 4 + 3 - 4 = 3$ , $∣ d ∣ = 21$ and $∣ n ∣ = 14$ .
Then $sin θ = \frac{3}{21 14} = \frac{3}{294} = \frac{6}{14} \approx 0.175$ .
$θ = arcsin \frac{6}{14} \approx 10. 1^{\circ}$ .

Answer: the ray makes an angle of $θ \approx 10. 1^{\circ}$ with the plane.

Exercise 90 BasicIB · P2 TZ1 May 2009

Calculate the angle between the lines $\frac{x - 1}{2} = 1 - y = 2 z$ and $x = y = 3 z$ .

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Answer: $θ = arccos \frac{7}{399} \approx 69. 5^{\circ}$

Show working and answer

First line: we call $s$ the common value, $\frac{x - 1}{2} = 1 - y = 2 z = s$ . Then $x = 1 + 2 s$ , $y = 1 - s$ , $z = \frac{s}{2}$ , and a direction vector is $d_{1} = 2 - 1 \frac{1}{2} ∥ 4 - 2 1$ .
Second line: $x = y = 3 z = u$ gives $x = u$ , $y = u$ , $z = \frac{u}{3}$ , with direction vector $d_{2} = 331$ .
The angle between two lines satisfies $cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣ ∣ d _{2} ∣}$ : $d_{1} \cdot d_{2} = 12 - 6 + 1 = 7, ∣ d_{1} ∣ = 21, ∣ d_{2} ∣ = 19 .$
$cos θ = \frac{7}{21 19} = \frac{7}{399} \approx 0.350$ , so $θ \approx 69. 5^{\circ}$ .

Answer: $θ = arccos \frac{7}{399} \approx 69. 5^{\circ}$

Exercise 91 HardIB · P2 TZ0 November 2012

Consider the planes $π_{1} : x - 2 y - 3 z = 2$ and $π_{2} : 2 x - y - z = k$ .

a) Calculate the angle between the planes $π_{1}$ and $π_{2}$ .

b) The planes $π_{1}$ and $π_{2}$ intersect in the line $L_{1}$ . Show that a vector equation of $L_{1}$ is $r = 0 2 - 3 k 2 k - 2 + t 15 - 3$ .

c) The line $L_{2}$ has Cartesian equation $5 - x = y + 3 = 2 - 2 z$ . The lines $L_{1}$ and $L_{2}$ intersect at a point $X$ . Calculate the coordinates of $X$ .

d) Determine a Cartesian equation of the plane $π_{3}$ that contains both lines $L_{1}$ and $L_{2}$ .

e) Let $Y$ be a point on $L_{1}$ and $Z$ a point on $L_{2}$ such that $X Y$ is perpendicular to $Y Z$ and the area of triangle $X Y Z$ is $3$ . Calculate the perimeter of triangle $X Y Z$ .

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Answer: a) $40. 2^{\circ}$ ; c) $X (1, 1, - 1)$ ; d) $x + 7 y + 12 z + 4 = 0$ ; e) perimeter $\approx 8.45$ . Part b) is proved.

Show working and answer

a) 1. The normal vectors are $n_{1} = (1, - 2, - 3)$ and $n_{2} = (2, - 1, - 1)$ . The angle between the planes is the acute angle between their normals. 2. $n_{1} \cdot n_{2} = 2 + 2 + 3 = 7$ , $∣ n_{1} ∣ = 14$ , $∣ n_{2} ∣ = 6$ . 3. $cos θ = \frac{7}{14 6} = \frac{7}{84}$ , so $θ \approx 40. 2^{\circ}$ .

b) 1. The point $(0, 2 - 3 k, 2 k - 2)$ lies on $π_{1}$ because $- 2 (2 - 3 k) - 3 (2 k - 2) = - 4 + 6 k - 6 k + 6 = 2$ ; and it lies on $π_{2}$ because $- (2 - 3 k) - (2 k - 2) = - 2 + 3 k - 2 k + 2 = k$ . 2. Direction: $d = (1, 5, - 3)$ satisfies $n_{1} \cdot d = 1 - 10 + 9 = 0$ and $n_{2} \cdot d = 2 - 5 + 3 = 0$ , so it is parallel to both planes. 3. A line that has a common point with both planes and is parallel to both is contained in both; since the planes are not parallel, that line is their intersection. This proves the result.

c) 1. Set $5 - x = y + 3 = 2 - 2 z = s$ . Then $x = 5 - s$ , $y = s - 3$ , $z = 1 - \frac{s}{2}$ . 2. A point on $L_{1}$ is $(t, 2 - 3 k + 5 t, 2 k - 2 - 3 t)$ . Equating it with the point on $L_{2}$ : $t = 5 - s, 2 - 3 k + 5 t = s - 3, 2 k - 2 - 3 t = 1 - \frac{s}{2} .$ 3. Substituting $t = 5 - s$ : $2 - 3 k + 25 - 5 s = s - 3$ , that is $3 k = 30 - 6 s$ , so $k = 10 - 2 s$ . And the third gives $2 k - 2 - 15 + 3 s = 1 - \frac{s}{2}$ , that is $2 k = 18 - \frac{7 s}{2}$ . 4. With $k = 10 - 2 s$ : $20 - 4 s = 18 - \frac{7 s}{2}$ , from which $2 = \frac{s}{2}$ and $s = 4$ . Then $k = 2$ and $t = 1$ . 5. $X = (5 - 4, 4 - 3, 1 - 2) = (1, 1, - 1)$ .

d) 1. Directions: $d_{1} = (1, 5, - 3)$ and $d_{2} = (- 1, 1, - \frac{1}{2})$ , or equivalently $(- 2, 2, - 1)$ . 2. Normal of $π_{3}$ : $d_{1} \times (- 2, 2, - 1) = 5 \cdot (- 1) - (- 3) \cdot 2 (- 3) (- 2) - 1 \cdot (- 1) 1 \cdot 2 - 5 \cdot (- 2) = 1712$ . 3. $π_{3}$ passes through $X (1, 1, - 1)$ : $x + 7 y + 12 z = 1 + 7 - 12 = - 4$ , that is $x + 7 y + 12 z + 4 = 0$ .

e) 1. The triangle is right-angled at $Y$ , with $Y$ on $L_{1}$ and $Z$ on $L_{2}$ . The angle at $X$ is the acute angle $θ$ between $L_{1}$ and $L_{2}$ . 2. $d_{1} \cdot (- 2, 2, - 1) = - 2 + 10 + 3 = 11$ , $∣ d_{1} ∣ = 35$ , $∣ (- 2, 2, - 1) ∣ = 3$ , so $cos θ = \frac{11}{3 35}$ , $θ \approx 51. 7^{\circ}$ , and $tan θ = \frac{194}{11} \approx 1.266$ . 3. Let $a = X Y$ . Then $Y Z = a tan θ$ and $X Z = \frac{a}{cos θ}$ . 4. Area: $\frac{1}{2} a \cdot a tan θ = 3$ , from which $a^{2} = \frac{6}{tan θ} = \frac{66}{194} \approx 4.739$ and $a \approx 2.177$ . 5. $Y Z \approx 2.177 \cdot 1.266 \approx 2.756$ and $X Z \approx \frac{2.177}{0.6198} \approx 3.512$ . 6. Perimeter $= 2.177 + 2.756 + 3.512 \approx 8.45$ .

Answer: a) $40. 2^{\circ}$ ; c) $X (1, 1, - 1)$ ; d) $x + 7 y + 12 z + 4 = 0$ ; e) perimeter $\approx 8.45$ . Part b) is proved.

Exercise 92 HardIB · P1 TZ1 May 2016

Two planes have equations $Π_{1} : 4 x + y + z = 8$ and $Π_{2} : 4 x + 3 y - z = 0$ .

a) Calculate the cosine of the angle between the two planes, giving your answer in the form $\frac{p}{q}$ , where $p, q \in Z$ .

Let $L$ be the line of intersection of the two planes.

b) (i) Show that $L$ has direction $- 1 22$ .

(ii) Show that the point $A (1, 0, 4)$ lies on both planes.

(iii) Write down a vector equation of $L$ .

$B$ is the point of $Π_{1}$ with coordinates $(a, b, 1)$ .

c) Given that the vector $A B$ is perpendicular to $L$ , calculate the value of $a$ and the value of $b$ .

d) Show that $A B = 32$ .

The point $P$ lies on $L$ and $A B P = 4 5^{\circ}$ .

e) Find the coordinates of the two possible positions of $P$ .

Show answer

Answer: a) $cos θ = \frac{9}{13}$ ; b) (iii) $r = 104 + t - 1 22$ ; c) $a = 1$ , $b = 3$ ; e) $P (1 - 2, 22, 4 + 22)$ or $P (1 + 2, - 22, 4 - 22)$ . Parts b) (i), (ii) and d) are proofs.

Show working and answer

a) 1. Normals: $n_{1} = (4, 1, 1)$ and $n_{2} = (4, 3, - 1)$ . 2. $n_{1} \cdot n_{2} = 16 + 3 - 1 = 18$ , $∣ n_{1} ∣ = 18$ , $∣ n_{2} ∣ = 26$ . 3. $cos θ = \frac{18}{18 26} = \frac{18}{26} = \frac{9}{13}$ .

b) (i) 1. The direction of $L$ is perpendicular to both normals, so it is $n_{1} \times n_{2}$ : $411 \times 43 - 1 = (1) (- 1) - (1) (3) (1) (4) - (4) (- 1) (4) (3) - (1) (4) = - 4 88 = 4 - 1 22 .$

(ii) 1. On $Π_{1}$ : $4 (1) + 0 + 4 = 8$ . On $Π_{2}$ : $4 (1) + 3 (0) - 4 = 0$ . Both equations hold.

(iii) 1. Using the point $A$ and the direction above: $L : r = 104 + t - 1 22$ , $t \in R$ .

c) 1. $B \in Π_{1}$ : $4 a + b + 1 = 8$ , i.e. $4 a + b = 7$ . 2. $A B = (a - 1, b, - 3)$ perpendicular to $L$ : $- (a - 1) + 2 b - 6 = 0$ , i.e. $- a + 2 b = 5$ . 3. From the second, $a = 2 b - 5$ . Substituting into the first: $4 (2 b - 5) + b = 7$ , $9 b = 27$ , $b = 3$ and $a = 1$ .

d) 1. With $a = 1$ and $b = 3$ : $A B = (0, 3, - 3)$ . 2. $A B = 0 + 9 + 9 = 18 = 32$ .

e) 1. $P$ lies on $L$ : $P = A + t (- 1, 2, 2)$ , so $A P = t (- 1, 2, 2)$ and $A P = 3 ∣ t ∣$ . 2. Since $A B ⊥ L$ , triangle $A B P$ is right-angled at $A$ . With $A B P = 4 5^{\circ}$ , it is isosceles: $A P = A B = 32$ . 3. So $3 ∣ t ∣ = 32$ , i.e. $t = \pm 2$ . 4. For $t = 2$ : $P = (1 - 2, 22, 4 + 22)$ . For $t = - 2$ : $P = (1 + 2, - 22, 4 - 22)$ .

Answer: a) $cos θ = \frac{9}{13}$ ; b) (iii) $r = 104 + t - 1 22$ ; c) $a = 1$ , $b = 3$ ; e) $P (1 - 2, 22, 4 + 22)$ or $P (1 + 2, - 22, 4 - 22)$ . Parts b) (i), (ii) and d) are proofs.

Exercise 93 HardIB · P1 TZ2 May 2016

A line $L$ has equation $\frac{x - 2}{p} = \frac{y - q}{2} = z - 1$ , where $p, q \in R$ . A plane $Π$ has equation $x + y + 3 z = 9$ .

a) Show that $L$ is not perpendicular to $Π$ .

b) Given that $L$ is contained in the plane $Π$ , find the value of $p$ and the value of $q$ .

Now consider a different case, in which $L$ and $Π$ form an acute angle $θ$ , where $θ = arcsin (\frac{1}{11})$ .

c) (i) Show that $p = - 2$ .

(ii) If $L$ and $Π$ intersect at $z = - 1$ , find the value of $q$ .

Show answer

Answer: b) $p = - 5$ , $q = 4$ ; c) (ii) $q = 10$ . Parts a) and c) (i) are proofs.

Show working and answer

a) 1. The direction of $L$ is $d = (p, 2, 1)$ and the normal of $Π$ is $n = (1, 1, 3)$ . 2. $L ⊥ Π$ would require $d$ and $n$ to be proportional. But $\frac{2}{1} \neq = \frac{1}{3}$ (second and third components), so they are not. Therefore $L$ is not perpendicular to $Π$ .

b) 1. $L$ contained in $Π$ requires $d \cdot n = 0$ : $p + 2 + 3 = 0$ , so $p = - 5$ . 2. Also the point $(2, q, 1)$ of $L$ must lie in $Π$ : $2 + q + 3 = 9$ , so $q = 4$ .

c) (i) 1. The angle between a line and a plane satisfies $sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣ ∣ n ∣}$ . With $d \cdot n = p + 5$ , $∣ d ∣ = p^{2} + 5$ and $∣ n ∣ = 11$ : $\frac{∣ p + 5 ∣}{p ^{2} + 5 11} = \frac{1}{11} \Rightarrow ∣ p + 5 ∣ = p^{2} + 5 .$ 2. Squaring: $p^{2} + 10 p + 25 = p^{2} + 5$ , so $10 p = - 20$ and $p = - 2$ .

(ii) 1. Let $z - 1 = μ$ be the common value of the fractions. With $z = - 1$ we get $μ = - 2$ . 2. Then $x = 2 + p μ = 2 + 4 = 6$ and $y = q + 2 μ = q - 4$ . 3. The point $(6, q - 4, - 1)$ lies on $Π$ : $6 + q - 4 - 3 = 9$ , so $q = 10$ .

Answer: b) $p = - 5$ , $q = 4$ ; c) (ii) $q = 10$ . Parts a) and c) (i) are proofs.

Exercise 94 HardIB · P1 TZ0 November 2005

The line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 1}{2}$ is reflected in the plane $x + y + z = 1$ .

Calculate the angle between the line and its image. Give your answer in radians.

Show answer

Answer: $θ = arccos (- \frac{23}{27}) \approx 2.59$ rad

Show working and answer

The line has direction $d = (1, 2, 2)$ and the plane has normal $n = (1, 1, 1)$ .
On reflecting in the plane, the component of $d$ along $n$ changes sign and the component parallel to the plane is preserved. The direction of the image is $d^{'} = d - 2 \frac{d \cdot n}{∣ n ∣ ^{2}} n = 122 - 2 \cdot \frac{5}{3} 111 = - 7/3 - 4/3 - 4/3 ∥ - 7 - 4 - 4 .$
We calculate the moduli and the dot product of $d$ and $(- 7, - 4, - 4)$ : $d \cdot d^{'} \to - 7 - 8 - 8 = - 23, ∣ d ∣ = 3, ∣ (- 7, - 4, - 4) ∣ = 49 + 16 + 16 = 9.$
Then $cos θ = \frac{- 23}{3 \cdot 9} = - \frac{23}{27} \Rightarrow θ = arccos (- \frac{23}{27}) \approx 2.59 rad .$
Check: if $φ$ is the angle between the line and the plane, $sin φ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣} = \frac{5}{3 3}$ , and the angle between the line and its image is $2 φ = 2 arcsin \frac{5}{3 3} \approx 2.59$ rad.

Answer: $θ = arccos (- \frac{23}{27}) \approx 2.59$ rad

Exercise 95 HardIB · P2 November 1998

The equations of the lines $L_{1}$ and $L_{2}$ are

$L_{1} : r = i - 20 j - 13 k + t (7 i + k)$ , where $t$ is a scalar;

$L_{2} : \frac{x + 30}{2} = \frac{y + 39}{1} = \frac{2 - 3 z}{2}$

The equations of the planes $P_{1}$ and $P_{2}$ are $P_{1} : 6 x + 3 y - 2 z = 12$

$P_{2} : r = 22 i + λ (i + j) + μ (i - 2 k)$ , where $λ$ and $μ$ are scalars.

a) Calculate the vector product $(i + j) \times (i - 2 k)$ .

b) i) Write down two vectors $n_{1}$ , $n_{2}$ perpendicular to the planes $P_{1}$ and $P_{2}$ respectively.

(ii) Using the above, or in any other way, calculate the acute angle between the planes, correct to one decimal place.

c) Show that $L_{2}$ is perpendicular to $P_{1}$ .

d) i) Calculate the coordinates of the point of intersection of $L_{2}$ and $P_{1}$ .

(ii) Using the above, or in any other way, show that the two lines and the two planes have a point in common.

Show answer

Answer: a) $- 2 i + 2 j - k$ ; b) $n_{1} = (6, 3, - 2)$ , $n_{2} = (- 2, 2, - 1)$ , angle $\approx 79. 0^{\circ}$ ; c) proved; d) $(8, - 20, - 12)$ , common point of $L_{1}$ , $L_{2}$ , $P_{1}$ and $P_{2}$ .

Show working and answer

a) 1. We expand with the components $(1, 1, 0)$ and $(1, 0, - 2)$ : $110 \times 10 - 2 = 1 \cdot (- 2) - 0 \cdot 0 0 \cdot 1 - 1 \cdot (- 2) 1 \cdot 0 - 1 \cdot 1 = - 2 2 - 1 = - 2 i + 2 j - k .$

b) i) 1. The normal of $P_{1} : 6 x + 3 y - 2 z = 12$ is read from the coefficients: $n_{1} = 63 - 2$ . 2. $P_{2}$ is generated by the vectors $i + j$ and $i - 2 k$ , so its normal is the cross product from part a): $n_{2} = - 2 2 - 1$ .

(ii) 1. The angle between the planes is the angle between their normals (taking the acute one): $cos α = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣} = \frac{∣ - 12 + 6 + 2∣}{49 9} = \frac{4}{21} .$ 2. $α = arccos \frac{4}{21} \approx 79. 0^{\circ}$ .

c) 1. The direction of $L_{2}$ is $(2, 1, - \frac{2}{3})$ , because $\frac{2 - 3 z}{2} = \frac{z - \frac{2}{3}}{- \frac{2}{3}}$ . Multiplying by 3 gives $(6, 3, - 2)$ . 2. This vector coincides with $n_{1}$ , the normal of $P_{1}$ . A line whose direction is the normal of a plane is perpendicular to that plane.

d) i) 1. Write $L_{2}$ with a parameter $s$ : $x = - 30 + 2 s$ , $y = - 39 + s$ , $z = \frac{2 - 2 s}{3}$ . 2. Substitute into $P_{1}$ : $6 (- 30 + 2 s) + 3 (- 39 + s) - 2 \cdot \frac{2 - 2 s}{3} = 12 \Rightarrow - 297 + 15 s - \frac{4}{3} + \frac{4 s}{3} = 12 \Rightarrow \frac{49}{3} s = \frac{931}{3} \Rightarrow s = 19.$ 3. The point is $x = - 30 + 38 = 8$ , $y = - 39 + 19 = - 20$ , $z = \frac{2 - 38}{3} = - 12$ : $(8, - 20, - 12)$ .

(ii) 1. We already know that $(8, - 20, - 12)$ lies on $L_{2}$ and on $P_{1}$ . 2. On $L_{1}$ : $x = 1 + 7 t = 8 \Rightarrow t = 1$ ; then $y = - 20$ and $z = - 13 + 1 = - 12$ . It holds, so it lies on $L_{1}$ . 3. On $P_{2}$ : $(8, - 20, - 12) = (22, 0, 0) + λ (1, 1, 0) + μ (1, 0, - 2)$ . From $z$ : $- 2 μ = - 12 \Rightarrow μ = 6$ ; from $y$ : $λ = - 20$ ; from $x$ : $22 - 20 + 6 = 8$ . It holds, so it lies on $P_{2}$ . 4. Therefore $(8, - 20, - 12)$ belongs to both lines and to both planes.

Distances

Exercise 96 MediumIB · P1 November 1999

Calculate the shortest distance from the point $A (0, 2, 2)$ to the line $r = 5 i + 9 j + 6 k + t (i + 2 j + 2 k)$ , where $t$ is a scalar.

Show answer

Answer: minimum distance $= 3$

Show working and answer

The line passes through $B (5, 9, 6)$ with direction vector $d = 122$ , whose magnitude is $∣ d ∣ = 1 + 4 + 4 = 3$ .
$A B = 574$ .
Cross product: $A B \times d = 7 \cdot 2 - 4 \cdot 2 4 \cdot 1 - 5 \cdot 2 5 \cdot 2 - 7 \cdot 1 = 6 - 6 3, A B \times d = 36 + 36 + 9 = 9.$
The distance from a point to a line is $\frac{A B \times d}{∣ d ∣} = \frac{9}{3} = 3$ .

Answer: minimum distance $= 3$

Exercise 97 MediumIB · P1 May 2000

Calculate the coordinates of the point on the line $L : x = 1 - λ, y = 2 - 3 λ, z = 2$ that is closest to the origin.

Show answer

Answer: $(\frac{3}{10}, - \frac{1}{10}, 2)$

Show working and answer

A general point on $L$ is $X (1 - λ, 2 - 3 λ, 2)$ .
The point closest to the origin is the one at which $O X$ is perpendicular to the line. The direction vector is $d = - 1 - 3 0$ .
$O X \cdot d = 0$ : $- (1 - λ) - 3 (2 - 3 λ) = 0$ , that is, $- 1 + λ - 6 + 9 λ = 0$ , so $10 λ = 7$ and $λ = \frac{7}{10}$ .
Substituting: $x = 1 - \frac{7}{10} = \frac{3}{10}$ , $y = 2 - \frac{21}{10} = - \frac{1}{10}$ , $z = 2$ .

Answer: $(\frac{3}{10}, - \frac{1}{10}, 2)$

Exercise 98 MediumIB · P1 May 2003

The point $A$ is the foot of the perpendicular from the point $(1, 1, 9)$ to the plane $2 x + y - z = 6$ . Find the coordinates of $A$ .

Show answer

Answer: $A (5, 3, 7)$

Show working and answer

The perpendicular to the plane from $Q (1, 1, 9)$ has as direction the normal of the plane, $n = (2, 1, - 1)$ : $(1 + 2 t, 1 + t, 9 - t)$ .
$A$ is the point of that line which lies in the plane: $2 (1 + 2 t) + (1 + t) - (9 - t) = 6$ .
We expand: $2 + 4 t + 1 + t - 9 + t = 6$ , that is, $6 t - 6 = 6$ , so $t = 2$ .
$A (1 + 4, 1 + 2, 9 - 2) = (5, 3, 7)$ .

Answer: $A (5, 3, 7)$

Exercise 99 MediumIB · P2 TZ0 November 2013

The line $L_{1}$ has equation $r = - 5 - 3 2 + λ - 1 22$ . The line $L_{2}$ passes through the origin, intersects $L_{1}$ and is perpendicular to $L_{1}$ .

a) Find a vector equation of $L_{2}$ .

b) Determine the shortest distance between the origin and $L_{1}$ .

Show answer

Answer: a) $r = t (- 14, - 11, 4)$ ; b) $d (O, L_{1}) = 37 \approx 6.08$

Show working and answer

a) 1. $L_{2}$ meets $L_{1}$ at a point $F = (- 5 - λ, - 3 + 2 λ, 2 + 2 λ)$ , and $O F$ is perpendicular to $L_{1}$ . 2. Perpendicularity condition with the direction vector $(- 1, 2, 2)$ : $(- 5 - λ) (- 1) + (- 3 + 2 λ) (2) + (2 + 2 λ) (2) = 0 \Rightarrow 5 + λ - 6 + 4 λ + 4 + 4 λ = 0 \Rightarrow 9 λ + 3 = 0.$ 3. $λ = - \frac{1}{3}$ and $F = (- \frac{14}{3}, - \frac{11}{3}, \frac{4}{3})$ . 4. $L_{2}$ passes through $O$ with direction $O F$ , which is proportional to $(- 14, - 11, 4)$ : $r = t - 14 - 11 4 or r = t - 14/3 - 11/3 4/3 .$

b) 1. The shortest distance from the origin to $L_{1}$ is $∣ O F ∣$ , since $F$ is the foot of the perpendicular. 2. $∣ O F ∣ = \frac{1}{3} 196 + 121 + 16 = \frac{333}{3} = \frac{333}{9} = 37$ .

Answer: a) $r = t (- 14, - 11, 4)$ ; b) $d (O, L_{1}) = 37 \approx 6.08$

Exercise 100 MediumIB · P1 TZ0 November 2004

a) Find the Cartesian equation of the plane that contains the origin O and the two points $A (1, 1, 1)$ and $B (2, - 1, 3)$ .

b) Find the distance from the point $C (10, 5, 1)$ to the plane $O A B$ .

Show answer

Answer: a) $4 x - y - 3 z = 0$ ; b) $d = \frac{32}{26} \approx 6.28$

Show working and answer

a) 1. A normal is $O A \times O B = 111 \times 2 - 1 3 = (1) (3) - (1) (- 1) (1) (2) - (1) (3) (1) (- 1) - (1) (2) = 4 - 1 - 3$ . 2. The plane is $4 x - y - 3 z = d$ and passes through the origin, so $d = 0$ : $4 x - y - 3 z = 0$ . Check with $B$ : $8 + 1 - 9 = 0$ .

b) 1. Distance from a point to a plane: $d = \frac{∣ 4 ( 10 ) - 5 - 3 ( 1 )∣}{16 + 1 + 9}$ . 2. $d = \frac{32}{26} \approx 6.28$ .

Answer: a) $4 x - y - 3 z = 0$ ; b) $d = \frac{32}{26} \approx 6.28$

Exercise 101 MediumIB · P1 TZ0 November 2006

Let $P$ be the point $(1, 0, - 2)$ and let $Π$ be the plane $x + y - 2 z + 3 = 0$ . Let $P^{'}$ be the point symmetric to $P$ with respect to the plane $Π$ . Find the coordinates of the point $P^{'}$ .

Show answer

Answer: $P^{'} (- \frac{5}{3}, - \frac{8}{3}, \frac{10}{3})$

Show working and answer

The normal to the plane is $n = (1, 1, - 2)$ . The line perpendicular to $Π$ through $P$ is $r = 10 - 2 + t 11 - 2 .$
We look for the foot of the perpendicular $M$ (the point of that line which lies in $Π$ ): $(1 + t) + t - 2 (- 2 - 2 t) + 3 = 0 \Rightarrow 1 + t + t + 4 + 4 t + 3 = 0 \Rightarrow 6 t + 8 = 0 \Rightarrow t = - \frac{4}{3} .$
Then $M = (1 - \frac{4}{3}, - \frac{4}{3}, - 2 + \frac{8}{3}) = (- \frac{1}{3}, - \frac{4}{3}, \frac{2}{3})$ .
$M$ is the midpoint of $P P^{'}$ , so $P^{'} = 2 M - P$ : $P^{'} = (- \frac{2}{3} - 1, - \frac{8}{3} - 0, \frac{4}{3} + 2) = (- \frac{5}{3}, - \frac{8}{3}, \frac{10}{3}) .$ (This is equivalent to using $t = - \frac{8}{3}$ , twice the parameter of the foot, in the line from step 1.)

Answer: $P^{'} (- \frac{5}{3}, - \frac{8}{3}, \frac{10}{3})$

Exercise 102 MediumIB · P2 TZ0 November 2006

a) The line $l_{1}$ passes through the point $A (0, 1, 2)$ and is perpendicular to the plane $x - 4 y - 3 z = 0$ .

Find the Cartesian equations of $l_{1}$ .

b) The line $l_{2}$ is parallel to $l_{1}$ and passes through the point $P (3, - 8, - 11)$ . Find the vector equation of the line $l_{2}$ .

c) i) The point $Q$ lies on the line $l_{1}$ such that $P Q$ is perpendicular to $l_{1}$ and $l_{2}$ .

Find the coordinates of $Q$ .

(ii) Hence, find the distance between $l_{1}$ and $l_{2}$ .

Show answer

Answer: a) $\frac{x}{1} = \frac{y - 1}{- 4} = \frac{z - 2}{- 3}$ ; b) $r = 3 - 8 - 11 + λ 1 - 4 - 3$ ; c) i) $Q (3, - 11, - 7)$ , ii) $d (l_{1}, l_{2}) = 5$

Show working and answer

a) 1. A line perpendicular to a plane has the normal of the plane as its direction: $d = (1, - 4, - 3)$ . 2. It passes through $A (0, 1, 2)$ : $l_{1} : \frac{x}{1} = \frac{y - 1}{- 4} = \frac{z - 2}{- 3} .$

b) 1. $l_{2}$ has the same direction and passes through $P (3, - 8, - 11)$ : $l_{2} : r = 3 - 8 - 11 + λ 1 - 4 - 3 .$

c) i) 1. A point on $l_{1}$ is $Q (μ, 1 - 4 μ, 2 - 3 μ)$ , so that $P Q = (μ - 3, 9 - 4 μ, 13 - 3 μ)$ . 2. Since $l_{1}$ and $l_{2}$ are parallel, $P Q$ being perpendicular to both is equivalent to $P Q \cdot (1, - 4, - 3) = 0$ : $(μ - 3) - 4 (9 - 4 μ) - 3 (13 - 3 μ) = μ - 3 - 36 + 16 μ - 39 + 9 μ = 26 μ - 78 = 0.$ 3. Hence $μ = 3$ and $Q = (3, 1 - 12, 2 - 9) = (3, - 11, - 7)$ .

(ii) 1. $P Q = Q - P = (0, - 3, 4)$ . 2. The distance between the parallel lines is $∣ P Q ∣ = 0 + 9 + 16 = 5$ .

Answer: a) $\frac{x}{1} = \frac{y - 1}{- 4} = \frac{z - 2}{- 3}$ ; b) $r = 3 - 8 - 11 + λ 1 - 4 - 3$ ; c) i) $Q (3, - 11, - 7)$ , ii) $d (l_{1}, l_{2}) = 5$

Exercise 103 MediumIB · P1 TZ1 May 2008

The points $A$ , $B$ , $C$ have position vectors $i + j + 2 k$ , $i + 2 j + 3 k$ , $3 i + k$ respectively and lie in the plane $π$ .

a) Calculate

(i) the area of triangle $A B C$ ;

(ii) the shortest distance from $C$ to the line $A B$ ;

(iii) the Cartesian equation of the plane $π$ .

The line $L$ passes through the origin, is normal to the plane $π$ , and meets $π$ at the point $D$ .

b) Calculate

(i) the coordinates of the point $D$ ;

(ii) the distance from $π$ to the origin.

Show answer

Answer: a) (i) $2$ ; (ii) $2$ ; (iii) $y - z + 1 = 0$ . b) (i) $D (0, - \frac{1}{2}, \frac{1}{2})$ ; (ii) $\frac{1}{2} = \frac{2}{2}$ .

Show working and answer

a) 1. The points are $A (1, 1, 2)$ , $B (1, 2, 3)$ , $C (3, 0, 1)$ . The side vectors are $A B = 011$ and $A C = 2 - 1 - 1$ . 2. Their vector (cross) product is $A B \times A C = 1 \cdot (- 1) - 1 \cdot (- 1) 1 \cdot 2 - 0 \cdot (- 1) 0 \cdot (- 1) - 1 \cdot 2 = 02 - 2, A B \times A C = 22 .$

(i) The area of the triangle is half the magnitude: $\frac{1}{2} \cdot 22 = 2$ .

(ii) 1. The area is also $\frac{1}{2} ∣ A B ∣ h$ , where $h$ is the distance from $C$ to the line $A B$ and $∣ A B ∣ = 0 + 1 + 1 = 2$ . 2. So $2 = \frac{1}{2} 2 h$ , hence $h = 2$ .

(iii) 1. The vector $02 - 2$ is normal to $π$ ; we use $01 - 1$ , so $π : y - z = d$ . 2. Through $A$ : $d = 1 - 2 = - 1$ . Hence $π : y - z = - 1$ , that is, $y - z + 1 = 0$ .

b) (i) 1. The line $L$ has the direction of the normal and passes through the origin: $(x, y, z) = (0, s, - s)$ . 2. For it to lie in $π$ : $s - (- s) + 1 = 0$ , that is, $2 s = - 1$ and $s = - \frac{1}{2}$ . 3. $D = (0, - \frac{1}{2}, \frac{1}{2})$ .

(ii) The distance from $π$ to the origin is $∣ O D ∣ = 0 + \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ . (Using the formula: $\frac{∣0 - 0 + 1∣}{0 + 1 + 1} = \frac{1}{2}$ .)

Answer: a) (i) $2$ ; (ii) $2$ ; (iii) $y - z + 1 = 0$ . b) (i) $D (0, - \frac{1}{2}, \frac{1}{2})$ ; (ii) $\frac{1}{2} = \frac{2}{2}$ .

Exercise 104 MediumIB · P1 TZ2 May 2021

The lines $L_{1}$ and $L_{2}$ are given by the following vector equations, where $λ, μ \in R$ .

$L_{1} : r_{1} = 32 - 1 + λ 2 - 2 2 L_{2} : r_{2} = 204 + μ 1 - 1 1$

a) Show that $L_{1}$ and $L_{2}$ do not intersect.

b) Find the minimum distance between $L_{1}$ and $L_{2}$ .

Show answer

Answer: a) the lines are parallel and distinct, so they do not intersect; b) $32$ ( $= 18 \approx 4.24$ ).

Show working and answer

a) 1. The direction vectors are $2 - 2 2 = 2 1 - 1 1$ : the lines are parallel. 2. For them to intersect they would have to coincide. The point $(3, 2, - 1)$ of $L_{1}$ lies on $L_{2}$ only if there is $μ$ with $2 + μ = 3$ , $- μ = 2$ and $4 + μ = - 1$ . The first gives $μ = 1$ and the second $μ = - 2$ : this is impossible. 3. They are parallel and distinct, so they do not intersect.

b) 1. The minimum distance between two parallel lines is the distance from a point of one of them to the other. We take $P (3, 2, - 1) \in L_{1}$ and $Q (2, 0, 4) \in L_{2}$ , with $d = (1, - 1, 1)$ . 2. $QP = 12 - 5$ and $QP \times d = 2 \cdot 1 - (- 5) (- 1) (- 5) (1) - 1 \cdot 1 1 \cdot (- 1) - 2 \cdot 1 = - 3 - 6 - 3, QP \times d = 54 = 36 .$ 3. The distance is $d = \frac{QP \times d}{∣ d ∣} = \frac{3 6}{3} = 32 \approx 4.24.$

Answer: a) the lines are parallel and distinct, so they do not intersect; b) $32$ ( $= 18 \approx 4.24$ ).

Exercise 105 HardIB · P2 May 2001

Triangle $A B C$ has vertices $A (- 1, 2, 3)$ , $B (- 1, 3, 5)$ and $C (0, - 1, 1)$ .

a) Calculate the value of the angle $θ$ between the vectors $A B$ and $A C$ .

b) Using part (a), or otherwise, calculate the area of triangle $A B C$ .

Let $L_{1}$ be the line parallel to $A B$ that passes through $D (2, - 1, 0)$ , and $L_{2}$ the line parallel to $A C$ that passes through $E (- 1, 1, 1)$ .

c) i) Find the equations of the lines $L_{1}$ and $L_{2}$ .

(ii) Show that $L_{1}$ and $L_{2}$ do not intersect.

d) Calculate the minimum distance between $L_{1}$ and $L_{2}$ .

Show answer

Answer: a) $θ \approx 146. 8^{\circ}$ ; b) Area $= \frac{21}{2} \approx 2.29$ ; c) i) $L_{1} : x = 2, y = - 1 + t, z = 2 t$ and $L_{2} : x = - 1 + s, y = 1 - 3 s, z = 1 - 2 s$ ; ii) they do not intersect; d) $\frac{9}{21} \approx 1.96$

Show working and answer

a) 1. $A B = 012$ and $A C = 1 - 3 - 2$ . 2. $A B \cdot A C = 0 - 3 - 4 = - 7$ , $∣ A B ∣ = 5$ , $∣ A C ∣ = 14$ . 3. $cos θ = \frac{- 7}{5 14} = \frac{- 7}{70}$ , so $θ \approx 146. 8^{\circ}$ .

b) 1. $A B \times A C = 1 \cdot (- 2) - 2 \cdot (- 3) 2 \cdot 1 - 0 \cdot (- 2) 0 \cdot (- 3) - 1 \cdot 1 = 42 - 1, A B \times A C = 21 .$ 2. Area $= \frac{1}{2} 21 \approx 2.29$ . (Alternatively: $\frac{1}{2} 70 sin θ$ , with the same result.)

c) i) 1. $L_{1}$ passes through $D (2, - 1, 0)$ with direction $A B = (0, 1, 2)$ : $x = 2$ , $y = - 1 + t$ , $z = 2 t$ . 2. $L_{2}$ passes through $E (- 1, 1, 1)$ with direction $A C = (1, - 3, - 2)$ : $x = - 1 + s$ , $y = 1 - 3 s$ , $z = 1 - 2 s$ .

(ii) 1. The direction vectors are not proportional, so the lines are not parallel. 2. If they intersected, there would be values $t$ and $s$ satisfying all three equations at once. From $x$ : $2 = - 1 + s$ , so $s = 3$ . 3. From $y$ : $- 1 + t = 1 - 9 = - 8$ , so $t = - 7$ . 4. From $z$ : with $t = - 7$ , $z = - 14$ , but with $s = 3$ , $z = 1 - 6 = - 5$ . They do not agree, so the lines do not intersect (they are skew).

d) 1. A vector perpendicular to both lines is $n = A B \times A C = 42 - 1$ . 2. $D E = - 3 21$ . 3. The distance between two skew lines is the projection of $D E$ onto $n$ : $d = \frac{∣ D E \cdot n ∣}{∣ n ∣} = \frac{∣ - 12 + 4 - 1∣}{21} = \frac{9}{21} \approx 1.96.$

Exercise 106 HardIB · P2 TZ0 November 2009

Two lines are defined as follows: $L_{1} : r = - 3 - 4 6 + λ 32 - 2$ and $L_{2} : \frac{x - 4}{- 3} = \frac{y + 7}{4} = - (z + 3)$ .

a) Find the coordinates of the point $A$ on $L_{1}$ and of the point $B$ on $L_{2}$ such that $A B$ is perpendicular to both $L_{1}$ and $L_{2}$ .

b) Find $∣ A B ∣$ .

c) Find the Cartesian equation of the plane $π$ which contains $L_{1}$ and does not intersect $L_{2}$ .

Show answer

Answer: a) $A (3, 0, 2)$ and $B (1, - 3, - 4)$ ; b) $∣ A B ∣ = 7$ ; c) $2 x + 3 y + 6 z - 18 = 0$

Show working and answer

a) 1. We write $L_{2}$ in parametric form: $x = 4 - 3 s$ , $y = - 7 + 4 s$ , $z = - 3 - s$ , with direction vector $(- 3, 4, - 1)$ . 2. A generic point of $L_{1}$ is $A = (- 3 + 3 λ, - 4 + 2 λ, 6 - 2 λ)$ and one of $L_{2}$ is $B = (4 - 3 s, - 7 + 4 s, - 3 - s)$ . Then $A B = 7 - 3 s - 3 λ - 3 + 4 s - 2 λ - 9 - s + 2 λ .$ 3. $A B ⊥ L_{1}$ : $A B \cdot (3, 2, - 2) = 0$ , that is, $33 + s - 17 λ = 0$ . 4. $A B ⊥ L_{2}$ : $A B \cdot (- 3, 4, - 1) = 0$ , that is, $- 24 + 26 s - λ = 0$ . 5. From the first, $s = 17 λ - 33$ . Into the second: $26 (17 λ - 33) - λ = 24$ , that is, $441 λ = 882$ , so $λ = 2$ and $s = 1$ . 6. $A = (3, 0, 2)$ and $B = (1, - 3, - 4)$ .

b) 1. $A B = (- 2, - 3, - 6)$ , so $∣ A B ∣ = 4 + 9 + 36 = 7$ .

c) 1. The plane contains $L_{1}$ and does not intersect $L_{2}$ , so it is parallel to $L_{2}$ . Its direction vectors are $(3, 2, - 2)$ and $(- 3, 4, - 1)$ , and a normal vector is $32 - 2 \times - 3 4 - 1 = 2 \cdot (- 1) - (- 2) \cdot 4 (- 2) (- 3) - 3 \cdot (- 1) 3 \cdot 4 - 2 \cdot (- 3) = 6918 ∥ 236 .$ 2. The plane is $2 x + 3 y + 6 z = d$ . With the point $A (3, 0, 2)$ of $L_{1}$ : $d = 6 + 12 = 18$ . 3. We check that $L_{2}$ does not lie in the plane: the point $(4, - 7, - 3)$ gives $8 - 21 - 18 = - 31 \neq = 18$ . Since the line is parallel to the plane and has no points on it, they do not intersect.

Answer: a) $A (3, 0, 2)$ and $B (1, - 3, - 4)$ ; b) $∣ A B ∣ = 7$ ; c) $2 x + 3 y + 6 z - 18 = 0$

Exercise 107 HardIB · P2 TZ0 November 2004

The equation of a line $l_{1}$ is $\frac{x + 2}{3} = \frac{y}{1} = \frac{z - 9}{- 2}$ .

a) Let $M$ be a point on $l_{1}$ with parameter $μ$ .

Express the coordinates of $M$ in terms of $μ$ .

b) The line $l_{2}$ is parallel to $l_{1}$ and passes through $P (4, 0, - 3)$ .

(i) Write down an equation for $l_{2}$ .

(ii) Express $P M$ in terms of $μ$ .

c) The vector $P M$ is perpendicular to $l_{1}$ .

(i) Find the value of $μ$ .

(ii) Find the distance between $l_{1}$ and $l_{2}$ .

d) The plane $π_{1}$ contains $l_{1}$ and $l_{2}$ .

Find an equation of $π_{1}$ , giving your answer in the form $A x + B y + C z = D$ .

e) The plane $π_{2}$ has equation $x - 5 y - z = - 11$ .

Verify that $l_{1}$ is the line of intersection of the planes $π_{1}$ and $π_{2}$ .

Show answer

Answer: a) $M (- 2 + 3 μ, μ, 9 - 2 μ)$ ; b) i) $l_{2} : r = 40 - 3 + t 31 - 2$ , ii) $P M = - 6 + 3 μ μ 12 - 2 μ$ ; c) i) $μ = 3$ , ii) $d = 36$ ; d) $2 x - 4 y + z = 5$ ; e) verified.

Show working and answer

a) 1. Setting each fraction equal to the parameter $μ$ : $x = - 2 + 3 μ$ , $y = μ$ , $z = 9 - 2 μ$ . 2. Hence, $M (- 2 + 3 μ, μ, 9 - 2 μ)$ .

b) i) 1. $l_{2}$ has the same direction $(3, 1, - 2)$ and passes through $P (4, 0, - 3)$ : $l_{2} : r = 40 - 3 + t 31 - 2, x = 4 + 3 t, y = t, z = - 3 - 2 t .$

(ii) 1. $P M = M - P = - 2 + 3 μ - 4 μ 9 - 2 μ + 3 = - 6 + 3 μ μ 12 - 2 μ$ .

c) i) 1. $P M ⊥ l_{1}$ means $P M \cdot (3, 1, - 2) = 0$ : $3 (- 6 + 3 μ) + μ - 2 (12 - 2 μ) = - 18 + 9 μ + μ - 24 + 4 μ = 14 μ - 42 = 0.$ 2. Hence $μ = 3$ .

(ii) 1. With $μ = 3$ : $P M = (3, 3, 6)$ . 2. Since it is perpendicular to both parallel lines, its magnitude is the distance between them: $d (l_{1}, l_{2}) = ∣ P M ∣ = 9 + 9 + 36 = 54 = 36 \approx 7.35.$

d) 1. The plane $π_{1}$ contains the direction $(3, 1, - 2)$ and the vector $P M = (3, 3, 6)$ , which is proportional to $(1, 1, 2)$ . 2. The normal is $31 - 2 \times 112 = 1 \cdot 2 - (- 2) \cdot 1 (- 2) \cdot 1 - 3 \cdot 2 3 \cdot 1 - 1 \cdot 1 = 4 - 8 2 ∥ 2 - 4 1 .$ 3. The plane is $2 x - 4 y + z = D$ . With $P (4, 0, - 3)$ : $D = 8 - 0 - 3 = 5$ . 4. $π_{1} : 2 x - 4 y + z = 5$ (equivalent to $- 2 x + 4 y - z + 5 = 0$ ).

e) 1. A point on $l_{1}$ , for example $(- 2, 0, 9)$ ( $μ = 0$ ), satisfies $π_{1}$ : $- 4 - 0 + 9 = 5$ ; and it satisfies $π_{2}$ : $- 2 - 0 - 9 = - 11$ . 2. The direction $(3, 1, - 2)$ satisfies both planes (it is perpendicular to their normals): for $π_{1}$ , $6 - 4 - 2 = 0$ ; for $π_{2}$ , $3 - 5 + 2 = 0$ . 3. Hence $l_{1}$ is contained in $π_{1}$ and in $π_{2}$ . Since the normals $(2, - 4, 1)$ and $(1, - 5, - 1)$ are not proportional, the planes are not parallel and meet in a single line. That line is $l_{1}$ .

Answer: a) $M (- 2 + 3 μ, μ, 9 - 2 μ)$ ; b) i) $l_{2} : r = 40 - 3 + t 31 - 2$ , ii) $P M = - 6 + 3 μ μ 12 - 2 μ$ ; c) i) $μ = 3$ , ii) $d = 36$ ; d) $2 x - 4 y + z = 5$ ; e) verified.

Exercise 108 HardIB · P1 TZ1 May 2011

The points $A (1, 2, 1)$ , $B (- 3, 1, 4)$ , $C (5, - 1, 2)$ and $D (5, 3, 7)$ are the vertices of a tetrahedron.

a) Find the vectors $A B$ and $A C$ .

b) Find the Cartesian equation of the plane $Π$ that contains the face $A B C$ .

c) Find the vector equation of the line that passes through $D$ and is perpendicular to $Π$ .

Using the above, or otherwise, calculate the shortest distance from $D$ to $Π$ .

d) (i) Calculate the area of triangle $A B C$ .

(ii) Calculate the volume of the tetrahedron $A B C D$ .

e) Determine which of the vertices $B$ or $D$ is closer to its opposite face.

Show answer

Answer: a) $A B = - 4 - 1 3$ , $A C = 4 - 3 1$ . b) $x + 2 y + 2 z - 7 = 0$ . c) $r = 537 + t 122$ ; $d (D, Π) = 6$ . d) (i) $12$ ; (ii) $24$ . e) $B$ is closer to its opposite face.

Show working and answer

a) $A B = B - A = - 4 - 1 3$ and $A C = C - A = 4 - 3 1$ .

b) 1. A normal to the plane is $A B \times A C = (- 1) \cdot 1 - 3 \cdot (- 3) 3 \cdot 4 - (- 4) \cdot 1 (- 4) (- 3) - (- 1) \cdot 4 = 81616 = 8122 .$ 2. Hence $Π : x + 2 y + 2 z = d$ , and with $A (1, 2, 1)$ : $d = 1 + 4 + 2 = 7$ . 3. $Π : x + 2 y + 2 z - 7 = 0$ .

c) 1. The line through $D$ perpendicular to $Π$ has the direction of the normal: $r = 537 + t 122$ . 2. The distance from $D$ to $Π$ is $d (D, Π) = \frac{∣5 + 2 \cdot 3 + 2 \cdot 7 - 7∣}{1 + 4 + 4} = \frac{18}{3} = 6.$

d) (i) The area of the triangle is half the magnitude of the vector (cross) product: $\frac{1}{2} 81616 = \frac{1}{2} \cdot 8 \cdot 3 = 12$ .

(ii) The volume is $\frac{1}{3} \cdot area of the base \cdot height = \frac{1}{3} \cdot 12 \cdot 6 = 24$ .

e) 1. The distance from $D$ to its opposite face $A B C$ is $6$ . 2. For $B$ we need to calculate the distance $h_{B}$ to the face $A C D$ . We use $V = \frac{1}{3} Area (A C D) h_{B}$ . 3. $C A = - 4 3 - 1$ and $C D = 045$ , with $C A \times C D = 1920 - 16$ and magnitude $361 + 400 + 256 = 1017$ . Hence $Area (A C D) = \frac{1}{2} 1017 \approx 15.9$ . 4. $h_{B} = \frac{3 V}{Area ( A C D )} = \frac{72}{1017} \approx 2.26$ . 5. Since $2.26 < 6$ , the vertex $B$ is closer to its opposite face.

Answer: a) $A B = - 4 - 1 3$ , $A C = 4 - 3 1$ . b) $x + 2 y + 2 z - 7 = 0$ . c) $r = 537 + t 122$ ; $d (D, Π) = 6$ . d) (i) $12$ ; (ii) $24$ . e) $B$ is closer to its opposite face.

Exercise 109 HardIB · P1 TZ1 May 2015

Two lines $l_{1}$ and $l_{2}$ are given respectively by the equations $r_{1} = O A + λ v$ and $r_{2} = O B + μ w$ , where $O A = i + 2 j + 3 k$ , $v = i + j + k$ , $O B = 2 i + j - k$ , $w = i - j + 2 k$ , and $O$ is the origin. Let $P$ be a point on $l_{1}$ and let $Q$ be a point on $l_{2}$ .

a) Calculate $P Q$ in terms of $λ$ and $μ$ .

b) Find the value of $λ$ and the value of $μ$ for which $P Q$ is perpendicular to the direction vectors of $l_{1}$ and $l_{2}$ .

c) Using the above, find the shortest distance between $l_{1}$ and $l_{2}$ .

d) Find the Cartesian equation of the plane $Π$ , which contains the line $l_{1}$ and is parallel to the direction vector of the line $l_{2}$ .

Let $O T = O B + η (v \times w)$ .

e) Find the value of $η$ for which the point $T$ lies in the plane $Π$ .

f) For this value of $η$ , calculate $B T$ .

g) State what you notice about your answers to (c) and (f), and give an interpretation of this result.

Show answer

Answer: a) $P Q = 1 + μ - λ - 1 - μ - λ - 4 + 2 μ - λ$ . b) $λ = - \frac{6}{7}$ , $μ = \frac{5}{7}$ . c) $\frac{6 14}{7}$ . d) $3 x - y - 2 z + 5 = 0$ . e) $η = - \frac{6}{7}$ . f) $\frac{6 14}{7}$ . g) Both coincide: it is the shortest distance between the two lines.

Show working and answer

a) 1. $P = A + λ v = (1 + λ, 2 + λ, 3 + λ)$ and $Q = B + μ w = (2 + μ, 1 - μ, - 1 + 2 μ)$ . 2. $P Q = Q - P = 1 + μ - λ - 1 - μ - λ - 4 + 2 μ - λ$ .

b) 1. $P Q \cdot v = 0$ : $(1 + μ - λ) + (- 1 - μ - λ) + (- 4 + 2 μ - λ) = - 4 + 2 μ - 3 λ = 0$ . 2. $P Q \cdot w = 0$ : $(1 + μ - λ) - (- 1 - μ - λ) + 2 (- 4 + 2 μ - λ) = - 6 + 6 μ - 2 λ = 0$ , that is, $λ = 3 μ - 3$ . 3. Substituting into the first: $- 4 + 2 μ - 9 μ + 9 = 0$ , so $5 - 7 μ = 0$ and $μ = \frac{5}{7}$ . 4. Then $λ = \frac{15}{7} - 3 = - \frac{6}{7}$ .

c) 1. With these values, $P Q = 1 + \frac{5}{7} + \frac{6}{7} - 1 - \frac{5}{7} + \frac{6}{7} - 4 + \frac{10}{7} + \frac{6}{7} = \frac{1}{7} 18 - 6 - 12 = \frac{6}{7} 3 - 1 - 2$ . 2. The shortest distance between the lines is the length of this common perpendicular segment: $P Q = \frac{6}{7} 9 + 1 + 4 = \frac{6 14}{7} \approx 3.21.$

d) 1. The normal of $Π$ is perpendicular to $v$ and to $w$ : $v \times w = 1 \cdot 2 - 1 \cdot (- 1) 1 \cdot 1 - 1 \cdot 2 1 \cdot (- 1) - 1 \cdot 1 = 3 - 1 - 2 .$ 2. Then $Π : 3 x - y - 2 z = d$ , and with $A (1, 2, 3) \in l_{1}$ : $d = 3 - 2 - 6 = - 5$ . 3. $Π : 3 x - y - 2 z + 5 = 0$ .

e) 1. $T = B + η (v \times w) = (2 + 3 η, 1 - η, - 1 - 2 η)$ . 2. Imposing this on $Π$ : $3 (2 + 3 η) - (1 - η) - 2 (- 1 - 2 η) + 5 = 0$ , that is, $6 + 9 η - 1 + η + 2 + 4 η + 5 = 12 + 14 η = 0$ . 3. $η = - \frac{6}{7}$ .

f) $B T = η (v \times w)$ , so $B T = ∣ η ∣ 9 + 1 + 4 = \frac{6}{7} 14 = \frac{6 14}{7}$ .

g) Both distances coincide: $P Q = B T = \frac{6 14}{7}$ . The vector $B T$ is perpendicular to $Π$ and to $l_{2}$ , so its length is the distance between the plane $Π$ and the line $l_{2}$ (which is parallel to $Π$ ), and this is the shortest distance between $l_{1}$ (contained in $Π$ ) and $l_{2}$ .

Relative positions and intersections

Exercise 110 BasicIB · P1 May 1999

Find the coordinates of the point where the line given by the parametric equations $x y z = 2 λ + 4 = - λ - 2 = 3 λ + 2$ intersects the plane with equation $2 x + 3 y - z = 2$ .

Show answer

Answer: point of intersection $(2, - 1, - 1)$

Show working and answer

We substitute the equations of the line into the equation of the plane: $2 (2 λ + 4) + 3 (- λ - 2) - (3 λ + 2) = 2.$
Expanding: $4 λ + 8 - 3 λ - 6 - 3 λ - 2 = 2$ , that is, $- 2 λ = 2$ , so $λ = - 1$ .
Returning to the parametric equations with $λ = - 1$ : $x = 2$ , $y = - 1$ , $z = - 1$ .

Answer: point of intersection $(2, - 1, - 1)$

Exercise 111 BasicIB · P1 TZ1 May 2004

The line $x - 1 = \frac{y + 1}{2} = \frac{z}{3}$ and the plane $r \cdot (i + 2 j - k) = 1$ intersect at the point $P$ . Calculate the coordinates of $P$ .

Show answer

Answer: $P (2, 1, 3)$

Show working and answer

We write the line in parametric form by equating the three fractions to $t$ : $x = 1 + t$ , $y = - 1 + 2 t$ , $z = 3 t$ .
The plane $r \cdot (i + 2 j - k) = 1$ in coordinates is $x + 2 y - z = 1$ .
We substitute the line into the plane: $(1 + t) + 2 (- 1 + 2 t) - 3 t = 1$ , that is, $2 t - 1 = 1$ , so $t = 1$ .
With $t = 1$ : $x = 2$ , $y = 1$ , $z = 3$ .

Answer: $P (2, 1, 3)$

Exercise 112 BasicIB · P1 TZ2 May 2004

The line $r = i + k + μ (i - j + 2 k)$ and the plane $2 x - y + z + 2 = 0$ intersect at the point $P$ . Find the coordinates of $P$ .

Show answer

Answer: $P (0, 1, - 1)$

Show working and answer

The parametric equations of the line are $x = 1 + μ$ , $y = - μ$ , $z = 1 + 2 μ$ .
We substitute into the plane: $2 (1 + μ) - (- μ) + (1 + 2 μ) + 2 = 0$ , that is, $5 μ + 5 = 0$ , so $μ = - 1$ .
With $μ = - 1$ : $x = 0$ , $y = 1$ , $z = - 1$ .

Answer: $P (0, 1, - 1)$

Exercise 113 BasicIB · P2 TZ0 November 2015

The three planes whose Cartesian equations are $2 x + 3 y - z = 11$ , $x + 2 y + z = 3$ and $5 x - y - z = 10$ intersect at the point $P$ . Find the coordinates of $P$ .

Show answer

Answer: $P (\frac{17}{9}, \frac{5}{3}, - \frac{20}{9})$

Show working and answer

Add the first two equations to eliminate $z$ : $3 x + 5 y = 14$ .
Add the second and third equations: $6 x + y = 13$ , that is $y = 13 - 6 x$ .
Substitute into $3 x + 5 y = 14$ : $3 x + 65 - 30 x = 14$ , so $- 27 x = - 51$ and $x = \frac{17}{9}$ .
$y = 13 - \frac{102}{9} = \frac{15}{9} = \frac{5}{3}$ .
From the second equation, $z = 3 - x - 2 y = 3 - \frac{17}{9} - \frac{30}{9} = - \frac{20}{9}$ .
In Paper 2 this is checked by solving the system with the calculator.

Answer: $P (\frac{17}{9}, \frac{5}{3}, - \frac{20}{9})$

Exercise 114 BasicIB · P1 TZ0 November 2016

Find the coordinates of the point of intersection of the planes defined by the equations $x + y + z = 3$ , $x - y + z = 5$ , $x + y + 2 z = 6$ .

Show answer

Answer: $(1, - 1, 3)$

Show working and answer

Subtract the first equation from the third: $(x + y + 2 z) - (x + y + z) = 6 - 3$ , so $z = 3$ .
Subtract the second equation from the first: $(x + y + z) - (x - y + z) = 3 - 5$ , so $2 y = - 2$ and $y = - 1$ .
Substitute into the first: $x - 1 + 3 = 3$ , so $x = 1$ .
Check in the second: $1 + 1 + 3 = 5$ , and in the third: $1 - 1 + 6 = 6$ . Both hold. (The $3 \times 3$ system can be solved directly with the GDC.)

Answer: $(1, - 1, 3)$

Exercise 115 BasicIB · P1 TZ2 May 2016

The following system of equations represents three planes in space.

$x + 3 y + z = - 1$

$x + 2 y - 2 z = 15$

$2 x + y - z = 6$

Find the coordinates of the point of intersection of the three planes.

Show answer

Answer: $(- 1, 2, - 6)$

Show working and answer

The point of intersection is the solution of the system. It can be solved with the GDC, or by elimination.
Subtract the second equation from the first: $y + 3 z = - 16$ . (1)
Subtract the third equation from twice the second: $2 (x + 2 y - 2 z) - (2 x + y - z) = 30 - 6$ , that is $3 y - 3 z = 24$ , or $y - z = 8$ . (2)
Subtract (2) from (1): $4 z = - 24$ , so $z = - 6$ .
From (2): $y = 8 + z = 2$ . From the first equation: $x = - 1 - 3 y - z = - 1 - 6 + 6 = - 1$ .
Check in the third: $2 (- 1) + 2 - (- 6) = 6$ . It holds.

Answer: $(- 1, 2, - 6)$

Exercise 116 BasicIB · P1 TZ1 May 2005

The line $\frac{x - 3}{2} = y + 1 = \frac{5 - z}{3}$ and the plane $2 x - y + 3 z = 10$ meet at the point $P$ .

Calculate the coordinates of $P$ .

Show answer

Answer: $P (7, 1, - 1)$

Show working and answer

Equating the three fractions to a parameter $s$ : $x = 3 + 2 s$ , $y = - 1 + s$ , $5 - z = 3 s \Rightarrow z = 5 - 3 s$ .
Substituting into the plane: $2 (3 + 2 s) - (- 1 + s) + 3 (5 - 3 s) = 6 + 4 s + 1 - s + 15 - 9 s = 22 - 6 s = 10.$
Solving: $6 s = 12$ , so $s = 2$ .
Back in the line: $x = 7$ , $y = 1$ , $z = - 1$ .

Answer: $P (7, 1, - 1)$

Exercise 117 BasicIB · P1 TZ0 November 2007Proof

A plane $Π$ has equation $r \cdot 2 - 1 1 = 16$ and a line $l$ has equations $\frac{x - 4}{- 1} = \frac{y + 2}{2} = \frac{z - 6}{4}$ .

Verify that the line $l$ is contained in the plane $Π$ .

View solution

The plane is $2 x - y + z = 16$ , with normal $n = (2, - 1, 1)$ . The line passes through $(4, - 2, 6)$ and has direction $d = (- 1, 2, 4)$ .
For the line to be contained in the plane, two things are needed: a point of the line must lie on the plane, and the direction must be perpendicular to the normal.
Point: $2 \cdot 4 - (- 2) + 6 = 8 + 2 + 6 = 16$ . The point $(4, - 2, 6)$ lies on $Π$ .
Direction: $d \cdot n = (- 1) (2) + 2 (- 1) + 4 \cdot 1 = - 2 - 2 + 4 = 0$ . The line is parallel to the plane.
A line parallel to a plane that has a point on it is contained in the plane.

This completes the proof.

Exercise 118 MediumIB · P2 TZ2 May 2010

The three planes $2 x - 2 y - z 4 x + 5 y - 2 z 3 x + 4 y - 3 z = 3 = - 3 = - 7$ intersect at a point with coordinates $(a, b, c)$ .

a) Find the values of $a$ , $b$ and $c$ .

b) The equations of three other planes are: $2 x - 4 y - 3 z - x + 3 y + 5 z 3 x - 5 y - z = 4 = - 2 = 6$ Find a vector equation of the line of intersection of these three planes.

Show answer

Answer: a) $a = 2$ , $b = - 1$ , $c = 3$ ; b) $r = 200 + t - 11 - 7 2$

Show working and answer

a) 1. We eliminate $x$ . Subtracting twice the first equation from the second: $(4 x + 5 y - 2 z) - 2 (2 x - 2 y - z) = - 3 - 6$ , that is, $9 y = - 9$ , so $y = - 1$ . 2. We also eliminate $x$ between the first and the third equations: $2 \cdot (3 x + 4 y - 3 z) - 3 \cdot (2 x - 2 y - z) = 2 \cdot (- 7) - 3 \cdot 3$ , that is, $14 y - 3 z = - 23$ . 3. With $y = - 1$ : $- 14 - 3 z = - 23$ , so $z = 3$ . 4. From the first equation: $2 x + 2 - 3 = 3$ , so $x = 2$ . 5. On Paper 2 this system can also be solved directly using the GDC.

b) 1. Adding twice the second equation to the first: $(2 x - 4 y - 3 z) + 2 (- x + 3 y + 5 z) = 4 + 2 (- 2)$ , that is, $2 y + 7 z = 0$ , so $y = - \frac{7 z}{2}$ . 2. From the second equation: $x = 3 y + 5 z + 2 = - \frac{21 z}{2} + 5 z + 2 = 2 - \frac{11 z}{2}$ . 3. Checking with the third: $3 x - 5 y - z = 6 - \frac{33 z}{2} + \frac{35 z}{2} - z = 6$ , which holds for all $z$ . The three planes share a line. 4. Taking $z = 2 t$ : $x = 2 - 11 t$ , $y = - 7 t$ , $z = 2 t$ .

Answer: a) $a = 2$ , $b = - 1$ , $c = 3$ ; b) $r = 200 + t - 11 - 7 2$

Exercise 119 MediumIB · P1 May 2001

Calculate the equation of the line of intersection of the two planes $- 4 x + y + z = - 2$ and $3 x - y + 2 z = - 1$ .

Show answer

Answer: $x = 3 + 3 t, y = 10 + 11 t, z = t$

Show working and answer

A direction vector of the line is perpendicular to both normal vectors $n_{1} = - 4 11$ and $n_{2} = 3 - 1 2$ : $d = n_{1} \times n_{2} = 1 \cdot 2 - 1 \cdot (- 1) 1 \cdot 3 - (- 4) \cdot 2 (- 4) (- 1) - 1 \cdot 3 = 3111 .$
We look for a common point. With $z = 0$ : $- 4 x + y = - 2$ and $3 x - y = - 1$ . Adding them, $- x = - 3$ , so $x = 3$ and $y = 10$ . The point is $(3, 10, 0)$ .
Check in both planes: $- 12 + 10 = - 2$ and $9 - 10 = - 1$ . Correct.
Taking $z = t$ gives parametric equations.

Answer: $x = 3 + 3 t, y = 10 + 11 t, z = t$

Exercise 120 MediumIB · P1 November 2002

Find an equation of the line of intersection of the following planes $x + 2 y - 3 z 2 x + 3 y - 5 z = 2 = 3$

Show answer

Answer: $\frac{x}{1} = \frac{y - 1}{1} = \frac{z}{1}$

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The direction vector is perpendicular to the two normal vectors $(1, 2, - 3)$ and $(2, 3, - 5)$ : $d = 2 \cdot (- 5) - (- 3) \cdot 3 (- 3) \cdot 2 - 1 \cdot (- 5) 1 \cdot 3 - 2 \cdot 2 = - 1 - 1 - 1,$ and we use the proportional vector $(1, 1, 1)$ .
We look for a common point. With $z = 0$ : $x + 2 y = 2$ and $2 x + 3 y = 3$ . From the first, $x = 2 - 2 y$ ; substituting into the second, $4 - 4 y + 3 y = 3$ , so $y = 1$ and $x = 0$ . The point is $(0, 1, 0)$ .
Check: $0 + 2 = 2$ and $0 + 3 = 3$ . Correct.
The line, in symmetric form, is $\frac{x}{1} = \frac{y - 1}{1} = \frac{z}{1}$ .

Answer: $\frac{x}{1} = \frac{y - 1}{1} = \frac{z}{1}$

Exercise 121 MediumIB · P1 May 2002

The vector equations of the lines $L_{1}$ and $L_{2}$ are

$L_{1} : r = i + j + k + λ (i + 2 j + 3 k)$

$L_{2} : r = i + 4 j + 5 k + μ (2 i + j + 2 k)$

The two lines intersect at the point $P$ . Find the position vector of $P$ .

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Answer: $O P = 357$

Show working and answer

We equate the coordinates of the points on the two lines: $(1 + λ, 1 + 2 λ, 1 + 3 λ) = (1 + 2 μ, 4 + μ, 5 + 2 μ)$ .
From the first component: $λ = 2 μ$ .
From the second: $1 + 4 μ = 4 + μ$ , so $μ = 1$ and $λ = 2$ .
Check in the third: $1 + 3 \cdot 2 = 7$ and $5 + 2 \cdot 1 = 7$ . It holds.
With $λ = 2$ : $P (3, 5, 7)$ .

Answer: $O P = 357$

Exercise 122 MediumIB · P1 TZ0 November 2010

Consider the plane with equation $4 x - 2 y - z = 1$ and the line given by the parametric equations

$x = 3 - 2 λ$

$y = (2 k - 1) + λ$

$z = - 1 + k λ$

Given that the line is perpendicular to the plane, find:

a) the value of $k$ .

b) the coordinates of the point of intersection of the line and the plane.

Show answer

Answer: (a) $k = \frac{1}{2}$ ; (b) $(\frac{5}{7}, \frac{8}{7}, - \frac{3}{7})$

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a) 1. The direction vector of the line is $(- 2, 1, k)$ and the normal vector of the plane is $(4, - 2, - 1)$ . 2. If the line is perpendicular to the plane, both vectors are parallel: $\frac{- 2}{4} = \frac{1}{- 2} = \frac{k}{- 1}$ . 3. The first two ratios equal $- \frac{1}{2}$ , so $\frac{k}{- 1} = - \frac{1}{2}$ and $k = \frac{1}{2}$ .

b) 1. With $k = \frac{1}{2}$ the line is $x = 3 - 2 λ$ , $y = λ$ , $z = - 1 + \frac{λ}{2}$ . 2. We substitute into the plane: $4 (3 - 2 λ) - 2 λ - (- 1 + \frac{λ}{2}) = 1$ , that is, $13 - \frac{21}{2} λ = 1$ , so $λ = \frac{8}{7}$ . 3. The point is $(3 - \frac{16}{7}, \frac{8}{7}, - 1 + \frac{4}{7}) = (\frac{5}{7}, \frac{8}{7}, - \frac{3}{7})$ .

Answer: (a) $k = \frac{1}{2}$ ; (b) $(\frac{5}{7}, \frac{8}{7}, - \frac{3}{7})$

Exercise 123 MediumIB · P1 TZ0 November 2004

Consider the four points $A (1, 4, - 1)$ , $B (2, 5, - 2)$ , $C (5, 6, 3)$ and $D (8, 8, 4)$ . Find the point of intersection of the lines $(A B)$ and $(C D)$ .

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Answer: $(- 1, 2, 1)$

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Direction vectors: $A B = (1, 1, - 1)$ and $C D = (3, 2, 1)$ .
Parametric equations: $(A B) : (1 + s, 4 + s, - 1 - s)$ and $(C D) : (5 + 3 t, 6 + 2 t, 3 + t)$ .
Equating the first two coordinates: $1 + s = 5 + 3 t$ and $4 + s = 6 + 2 t$ . Subtracting: $3 = 1 - t$ , so $t = - 2$ , and then $s = - 2$ .
We check the third coordinate: $- 1 - s = 1$ and $3 + t = 1$ . They agree, so the lines intersect.
The point is obtained with $s = - 2$ : $(1 - 2, 4 - 2, - 1 + 2) = (- 1, 2, 1)$ .

Answer: $(- 1, 2, 1)$

Exercise 124 MediumIB · P1 TZ0 November 2016

Consider the lines $l_{1}$ and $l_{2}$ defined by

$l_{1} : r = - 3 - 2 a + β 142$ and $l_{2} : \frac{6 - x}{3} = \frac{y - 2}{4} = 1 - z$ , where $a$ is a constant.

Given that the lines $l_{1}$ and $l_{2}$ intersect at a point P,

a) find the value of $a$ ;

b) determine the coordinates of the point of intersection P.

Show answer

Answer: a) $a = - 7$ ; b) $P (0, 10, - 1)$

Show working and answer

a) 1. Write $l_{2}$ in parametric form with $\frac{6 - x}{3} = \frac{y - 2}{4} = 1 - z = μ$ : $x = 6 - 3 μ$ , $y = 2 + 4 μ$ , $z = 1 - μ$ . 2. A point of $l_{1}$ is $(- 3 + β, - 2 + 4 β, a + 2 β)$ . Equating coordinates: $- 3 + β = 6 - 3 μ, - 2 + 4 β = 2 + 4 μ, a + 2 β = 1 - μ .$ 3. From the second: $β = 1 + μ$ . In the first: $- 3 + 1 + μ = 6 - 3 μ$ , $4 μ = 8$ , $μ = 2$ and $β = 3$ . 4. In the third: $a + 6 = 1 - 2 = - 1$ , so $a = - 7$ .

b) 1. With $μ = 2$ in $l_{2}$ : $x = 6 - 6 = 0$ , $y = 2 + 8 = 10$ , $z = 1 - 2 = - 1$ . 2. We check with $l_{1}$ ( $β = 3$ ): $(- 3 + 3, - 2 + 12, - 7 + 6) = (0, 10, - 1)$ .

Answer: a) $a = - 7$ ; b) $P (0, 10, - 1)$

Exercise 125 MediumIB · P2 November 1997

Consider the lines $L_{1}$ and $L_{2}$ given by the equations

$L_{1} : \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z}{4}$ $L_{2} : \frac{x + 2}{2} = y - 1 = 2 - z$

a) Show that the vector $- 2 i + 11 j + 7 k$ is perpendicular to the lines $L_{1}$ and $L_{2}$ .

b) Show that $L_{1}$ and $L_{2}$ do not intersect.

c) Find the equation of the plane that contains $L_{1}$ and is parallel to $L_{2}$ .

Give your answer in the form $a x + b y + cz = d$ .

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Answer: c) $- 2 x + 11 y + 7 z = 7$ . Parts a) and b) are proofs.

Show working and answer

a) 1. Direction of $L_{1}$ : equating the fractions to $s$ , $x = 2 + 3 s$ , $y = 1 - 2 s$ , $z = 4 s$ , so $d_{1} = (3, - 2, 4)$ and it passes through $(2, 1, 0)$ . 2. Direction of $L_{2}$ : equating to $t$ , $x = - 2 + 2 t$ , $y = 1 + t$ , $z = 2 - t$ , so $d_{2} = (2, 1, - 1)$ and it passes through $(- 2, 1, 2)$ . 3. $(- 2, 11, 7) \cdot (3, - 2, 4) = - 6 - 22 + 28 = 0$ and $(- 2, 11, 7) \cdot (2, 1, - 1) = - 4 + 11 - 7 = 0$ . It is perpendicular to both lines.

b) 1. The directions $(3, - 2, 4)$ and $(2, 1, - 1)$ are not proportional, so the lines are not parallel. 2. We look for a common point: $2 + 3 s = - 2 + 2 t$ , $1 - 2 s = 1 + t$ , $4 s = 2 - t$ . 3. From the second, $t = - 2 s$ . In the first: $2 + 3 s = - 2 - 4 s$ , so $s = - \frac{4}{7}$ and $t = \frac{8}{7}$ . 4. In the third: $4 s = - \frac{16}{7}$ and $2 - t = \frac{6}{7}$ . They do not match. 5. The system has no solution: the lines do not intersect.

c) 1. The plane contains $L_{1}$ and is parallel to $L_{2}$ , so its direction vectors are $d_{1}$ and $d_{2}$ . A normal is $d_{1} \times d_{2}$ , and by part a) we can take $n = (- 2, 11, 7)$ . 2. The plane is $- 2 x + 11 y + 7 z = d$ . Passing through $(2, 1, 0)$ of $L_{1}$ : $d = - 4 + 11 = 7$ . 3. $- 2 x + 11 y + 7 z = 7$ . Check with another point of $L_{1}$ , $(5, - 1, 4)$ : $- 10 - 11 + 28 = 7$ .

Answer: c) $- 2 x + 11 y + 7 z = 7$ . Parts a) and b) are proofs.

Exercise 126 MediumIB · P1 TZ0 November 2007

The lines $l_{1}$ and $l_{2}$ have equations $r_{1} = 430 + λ 15 - 2$ and $r_{2} = 2 - 1 3 + μ 02 - 3$ respectively, where $λ$ and $μ$ are parameters.

a) Verify that $l_{1}$ passes through the point $(2, - 7, 4)$ .

b) Determine whether the lines $l_{1}$ and $l_{2}$ intersect.

Show answer

Answer: a) $λ = - 2$ gives the point $(2, - 7, 4)$ . b) The lines do not intersect.

Show working and answer

a) 1. A point lies on $l_{1}$ if there exists a $λ$ that reproduces its three coordinates: $4 + λ = 2, 3 + 5 λ = - 7, - 2 λ = 4.$ 2. The first gives $λ = - 2$ , and this value also satisfies the second ( $3 - 10 = - 7$ ) and the third ( $- 2 \cdot (- 2) = 4$ ). 3. Hence $(2, - 7, 4)$ lies on $l_{1}$ .

b) 1. We equate the coordinates of a point on $l_{1}$ and a point on $l_{2}$ : $4 + λ 3 + 5 λ - 2 λ = 2 = - 1 + 2 μ = 3 - 3 μ$ 2. From the first, $λ = - 2$ . Substituting into the second: $3 - 10 = - 1 + 2 μ$ , so $μ = - 3$ . 3. We check in the third: the left-hand side equals $- 2 \cdot (- 2) = 4$ and the right-hand side equals $3 - 3 \cdot (- 3) = 12$ . They do not match. 4. The system has no solution, so the lines have no common point.

Answer: a) $λ = - 2$ gives the point $(2, - 7, 4)$ . b) The lines do not intersect.

Exercise 127 MediumIB · P2 TZ1 May 2008

Find the vector equation of the line of intersection of the three planes represented by the following system of equations.

$2 x - 7 y + 5 z 6 x + 3 y - z - 14 x - 23 y + 13 z = 1 = - 1 = 5$

Show answer

Answer: $r = - \frac{1}{12} - \frac{1}{6} 0 + t - 1 46$

Show working and answer

We use $z$ as the parameter and solve for $x$ and $y$ using the first two equations. From the second, $z = 6 x + 3 y + 1$ .
Substituting into the first: $2 x - 7 y + 5 (6 x + 3 y + 1) = 1$ , that is, $32 x + 8 y = - 4$ , i.e. $8 x + 2 y = - 1$ .
Substituting into the third: $- 14 x - 23 y + 13 (6 x + 3 y + 1) = 5$ , that is, $64 x + 16 y = - 8$ , i.e. $8 x + 2 y = - 1$ . This is the same equation, so the third equation adds no new information and the three planes meet in a line (the system is consistent and indeterminate).
Taking $x = s$ : $y = \frac{- 1 - 8 s}{2} = - \frac{1}{2} - 4 s$ and $z = 6 s + 3 (- \frac{1}{2} - 4 s) + 1 = - \frac{1}{2} - 6 s$ . This is a valid form; to get a more convenient point and vector, we reparametrise with $z = t$ .
From $z = - \frac{1}{2} - 6 s$ we obtain $s = - \frac{1}{12} - \frac{t}{6}$ , and so $x = - \frac{1}{12} - \frac{t}{6}$ , $y = - \frac{1}{6} + \frac{2}{3} t$ . Thus, with $z = t$ : $x = - \frac{1}{12} - \frac{1}{6} t, y = - \frac{1}{6} + \frac{2}{3} t, z = t .$
Multiplying the direction vector by $6$ (changing the parameter to $t = 6 u$ ): point $(- \frac{1}{12}, - \frac{1}{6}, 0)$ and direction $- 1 46$ .
Checking the direction vector: it is perpendicular to the three normals, since $2 (- 1) - 7 \cdot 4 + 5 \cdot 6 = 0$ , $6 (- 1) + 3 \cdot 4 - 6 = 0$ and $- 14 (- 1) - 23 \cdot 4 + 13 \cdot 6 = 0$ .

Answer: $r = - \frac{1}{12} - \frac{1}{6} 0 + t - 1 46$

Exercise 128 MediumIB · P1 TZ1 May 2012

The planes $2 x + 3 y - z = 5$ and $x - y + 2 z = k$ intersect in the line $5 x + 1 = 9 - 5 y = - 5 z$ .

Find the value of $k$ .

Show answer

Answer: $k = - 2$

Show working and answer

All points of the line must lie on the second plane. To obtain a point on the line we take $x = 0$ : $5 x + 1 = 1$ , so $9 - 5 y = 1$ gives $y = \frac{8}{5}$ and $- 5 z = 1$ gives $z = - \frac{1}{5}$ . The point is $(0, \frac{8}{5}, - \frac{1}{5})$ .
We check that it lies on the first plane: $2 \cdot 0 + 3 \cdot \frac{8}{5} - (- \frac{1}{5}) = \frac{24}{5} + \frac{1}{5} = 5$ . This holds.
We require it to lie on the second plane: $k = 0 - \frac{8}{5} + 2 (- \frac{1}{5}) = - \frac{10}{5} = - 2$ .
Check with another point: with $5 x + 1 = 9 - 5 y = - 5 z = u$ we have $x = \frac{u - 1}{5}$ , $y = \frac{9 - u}{5}$ , $z = - \frac{u}{5}$ , and so $x - y + 2 z = \frac{( u - 1 ) - ( 9 - u ) - 2 u}{5} = - 2$ for any $u$ .

Answer: $k = - 2$

Exercise 129 MediumIB · P2 TZ0 November 2011

Two planes $Π_{1}$ and $Π_{2}$ have equations $2 x + y + z = 1$ and $3 x + y - z = 2$ , respectively.

a) Find the vector equation of $L$ , the line of intersection of $Π_{1}$ and $Π_{2}$ .

b) Verify that the plane $Π_{3}$ , which is perpendicular to $Π_{1}$ and contains $L$ , has equation $x - 2 z = 1$ .

c) The point $P$ has coordinates $(- 2, 4, 1)$ , the point $Q$ lies on $Π_{3}$ and $P Q$ is perpendicular to $Π_{2}$ . Find the coordinates of $Q$ .

Show answer

Answer: a) $r = 1 - 1 0 + t 2 - 5 1$ . b) $Π_{3} : x - 2 z = 1$ . c) $Q (1, 5, 0)$ .

Show working and answer

a) 1. The direction of $L$ is perpendicular to both normals $n_{1} = 211$ and $n_{2} = 31 - 1$ : $n_{1} \times n_{2} = 1 \cdot (- 1) - 1 \cdot 1 1 \cdot 3 - 2 \cdot (- 1) 2 \cdot 1 - 1 \cdot 3 = - 2 5 - 1 ∥ 2 - 5 1 .$ 2. A point on $L$ : with $z = 0$ the system $2 x + y = 1$ , $3 x + y = 2$ gives (subtracting) $x = 1$ and $y = - 1$ . The point is $(1, - 1, 0)$ . 3. $L : r = 1 - 1 0 + t 2 - 5 1$ .

b) 1. $Π_{3}$ contains $L$ and is perpendicular to $Π_{1}$ , so its normal is perpendicular to the direction of $L$ and to $n_{1}$ . A suitable normal is $2 - 5 1 \times 211 = (- 5) \cdot 1 - 1 \cdot 1 1 \cdot 2 - 2 \cdot 1 2 \cdot 1 - (- 5) \cdot 2 = - 6 012 ∥ 10 - 2 .$ 2. Then $Π_{3} : x - 2 z = d$ . It passes through $(1, - 1, 0)$ , which lies on $L$ : $d = 1$ . 3. Therefore $Π_{3} : x - 2 z = 1$ , as required.

c) 1. $P Q$ is perpendicular to $Π_{2}$ , so $Q = P + s n_{2} = (- 2 + 3 s, 4 + s, 1 - s)$ . 2. $Q$ lies on $Π_{3}$ : $(- 2 + 3 s) - 2 (1 - s) = 1$ , that is, $- 4 + 5 s = 1$ and $s = 1$ . 3. $Q = (1, 5, 0)$ .

Answer: a) $r = 1 - 1 0 + t 2 - 5 1$ . b) $Π_{3} : x - 2 z = 1$ . c) $Q (1, 5, 0)$ .

Exercise 130 MediumIB · P1 TZ0 November 2012

Consider the following equations, where $a, b \in R$ :

$x + 3 y + (a - 1) z 2 x + 2 y + (a - 2) z 3 x + y + (a - 3) z = 1 = 1 = b$

a) If each of these equations defines a plane, show that, for any value of $a$ , the planes do not intersect at a unique point.

b) Find the value of $b$ for which the intersection of the planes is a line.

Show answer

Answer: a) The third row is $2 F_{2} - F_{1}$ , so the system has no unique solution. b) $b = 1$ .

Show working and answer

a) 1. The coefficients of the unknowns form the rows $F_{1} = (1, 3, a - 1)$ , $F_{2} = (2, 2, a - 2)$ and $F_{3} = (3, 1, a - 3)$ . 2. We have $2 F_{2} - F_{1} = (4 - 1, 4 - 3, 2 a - 4 - a + 1) = (3, 1, a - 3) = F_{3}$ for all $a$ . 3. Therefore the determinant of the coefficient matrix is $0$ for any $a$ (one row is a combination of the others), and the system never has a unique solution: it has either no solution or infinitely many.

b) 1. We apply the same combination to the constant terms: $2 \cdot 1 - 1 = 1$ . For the third equation to follow from the first two, we need $b = 1$ . 2. If $b \neq = 1$ , the third equation contradicts the other two and the system has no solution (the planes have no common point). 3. If $b = 1$ , the third equation is redundant and two independent equations remain (the planes $F_{1}$ and $F_{2}$ are not parallel, because $(1, 3)$ and $(2, 2)$ are not proportional). Two non-parallel planes intersect in a line.

Answer: a) The third row is $2 F_{2} - F_{1}$ , so the system has no unique solution. b) $b = 1$ .

Exercise 131 MediumIB · P1 TZ2 May 2014

a) Show that the following system of equations has infinitely many solutions.

$x + y + 2 z 3 x - y + 14 z x + 2 y = - 2 = 6 = - 5$

The system of equations represents three planes in space.

b) Find the parametric equations of the line of intersection of the three planes.

Show answer

Answer: a) The third equation is satisfied automatically once the other two are combined: infinitely many solutions. b) $x = 1 - 4 t, y = - 3 + 2 t, z = t$ .

Show working and answer

a) 1. Eliminate $x$ from the second and third equations using the first: $F_{2} - 3 F_{1}$ and $F_{3} - F_{1}$ : $x + y + 2 z - 4 y + 8 z y - 2 z = - 2 = 12 = - 3$ 2. Multiply the third by $4$ and add it to the second ( $4 F_{3} + F_{2}$ ): $x + y + 2 z - 4 y + 8 z 0 = - 2 = 12 = 0$ 3. The last equation is always satisfied, so two independent equations remain in three unknowns: the system is consistent and indeterminate, with infinitely many solutions.

b) 1. From $- 4 y + 8 z = 12$ we obtain $y = - 3 + 2 z$ . 2. Take $z = t$ : $y = - 3 + 2 t$ . From the first equation, $x = - 2 - y - 2 z = - 2 + 3 - 2 t - 2 t = 1 - 4 t$ . 3. Parametric equations: $x = 1 - 4 t, y = - 3 + 2 t, z = t .$ 4. Check in the third original equation: $x + 2 y = 1 - 4 t - 6 + 4 t = - 5$ .

Answer: a) The third equation is satisfied automatically once the other two are combined: infinitely many solutions. b) $x = 1 - 4 t, y = - 3 + 2 t, z = t$ .

Exercise 132 MediumIB · P2 TZ1 May 2009

Consider the planes defined by the equations $x + y + 2 z = 2$ , $2 x - y + 3 z = 2$ and $5 x - y + a z = 5$ , where $a$ is a real number.

a) If $a = 4$ , find the coordinates of the point of intersection of the three planes.

b) (i) Find the value of $a$ for which the planes do not intersect at a unique point.

(ii) For this value of $a$ , show that the three planes have no common point.

Show answer

Answer: a) $(\frac{11}{12}, \frac{7}{12}, \frac{1}{4})$ ; b) i) $a = 8$ ; ii) the system is inconsistent, so there is no common point.

Show working and answer

a) 1. Add the first two equations: $3 x + 5 z = 4$ . 2. Subtract the second from the third (with $a = 4$ ): $(5 x - y + 4 z) - (2 x - y + 3 z) = 5 - 2$ , that is, $3 x + z = 3$ . 3. Subtract this last equation from $3 x + 5 z = 4$ : $4 z = 1$ , so $z = \frac{1}{4}$ . 4. From $3 x + z = 3$ : $3 x = \frac{11}{4}$ , so $x = \frac{11}{12}$ . 5. From the first equation: $y = 2 - x - 2 z = 2 - \frac{11}{12} - \frac{1}{2} = \frac{7}{12}$ . 6. The point of intersection is $(\frac{11}{12}, \frac{7}{12}, \frac{1}{4})$ .

b) (i) 1. In general, subtracting the second equation from the third gives $3 x + (a - 3) z = 3$ . Together with $3 x + 5 z = 4$ , subtracting gives $(a - 8) z = - 1$ . 2. If $a \neq = 8$ , $z$ is determined and the system has a unique solution. The planes do not intersect at a unique point when $a = 8$ .

(Equivalently, the determinant of the coefficients is $24 - 3 a$ , which vanishes for $a = 8$ .)

(ii) 1. With $a = 8$ the equations $3 x + 5 z = 4$ and $3 x + 5 z = 3$ (the latter comes from subtracting the second equation from the third) contradict each other: $4 = 3$ is impossible. 2. The system is inconsistent, so the three planes have no common point.

Answer: a) $(\frac{11}{12}, \frac{7}{12}, \frac{1}{4})$ ; b) i) $a = 8$ ; ii) the system is inconsistent, so there is no common point.

Exercise 133 MediumIB · P2 TZ0 November 2008

a) Find the set of values of $k$ for which the following system of equations has no solution.

$x + 2 y - 3 z 3 x + y + 2 z 5 x + 7 z = k = 4 = 5$

b) Describe the geometrical relationship between the three planes represented by this system of equations.

Show answer

Answer: a) $k \neq = 3$ (for all $k \in R$ except $k = 3$ ); b) if $k = 3$ the three planes intersect in a line; if $k \neq = 3$ they have no common point and meet pairwise in parallel lines.

Show working and answer

a) 1. We look for whether the third equation is a linear combination of the first two on the left-hand side: $α (1, 2, - 3) + β (3, 1, 2) = (5, 0, 7)$ . From the $y$ -coordinate: $2 α + β = 0$ , so $β = - 2 α$ . From the $x$ -coordinate: $α + 3 β = 5$ , that is $- 5 α = 5$ , so $α = - 1$ and $β = 2$ . Check with $z$ : $3 + 4 = 7$ . Correct. 2. Therefore the determinant of the coefficients is $0$ , and the third equation is equivalent to $- (1st) + 2 (2nd)$ on the left-hand side. For the system to be consistent, the right-hand side must satisfy the same relation: $- k + 2 \cdot 4 = 5$ , that is $k = 3$ . 3. If $k \neq = 3$ , the equation $5 x + 7 z = 5$ contradicts the other two ( $5 x + 7 z = 8 - k$ ) and the system has no solution.

b) 1. For $k = 3$ : the equations have rank $2$ and no two planes are parallel. The three planes pass through a common line (they intersect in a line). 2. For $k \neq = 3$ : there is no point common to the three, but the planes are not parallel to each other. They meet pairwise in three parallel lines, forming a triangular prism.

Answer: a) $k \neq = 3$ (for all $k \in R$ except $k = 3$ ); b) if $k = 3$ the three planes intersect in a line; if $k \neq = 3$ they have no common point and meet pairwise in parallel lines.

Exercise 134 MediumIB · P1 TZ1 May 2022

Consider the three planes

$Π_{1} : 2 x - y + z = 4$

$Π_{2} : x - 2 y + 3 z = 5$

$Π_{3} : - 9 x + 3 y - 2 z = 32$

a) Show that the three planes do not intersect.

b) i) Verify that the point $P (1, - 2, 0)$ lies on $Π_{1}$ and $Π_{2}$ .

(ii) Calculate the vector equation of $L$ , the line of intersection of $Π_{1}$ and $Π_{2}$ .

c) Calculate the distance between $L$ and $Π_{3}$ .

Show answer

Answer: a) the system is inconsistent; b) ii) $r = 1 - 2 0 + λ 153$ ; c) $d = \frac{94}{2} \approx 4.85$ .

Show working and answer

a) 1. We look for a combination of $Π_{1}$ and $Π_{2}$ that reproduces the left-hand side of $Π_{3}$ : $- 5 (2 x - y + z) + (x - 2 y + 3 z) = - 9 x + 3 y - 2 z$ . 2. Applying the same combination to the right-hand sides: $- 5 \cdot 4 + 5 = - 15$ . Any point common to $Π_{1}$ and $Π_{2}$ would therefore satisfy $- 9 x + 3 y - 2 z = - 15$ . 3. But $Π_{3}$ requires $- 9 x + 3 y - 2 z = 32$ , and $- 15 \neq = 32$ . The system is inconsistent: the three planes have no common point.

b) i) $Π_{1}$ : $2 \cdot 1 - (- 2) + 0 = 4$ . Correct. $Π_{2}$ : $1 - 2 (- 2) + 3 \cdot 0 = 5$ . Correct.

(ii) 1. The direction vector of $L$ is perpendicular to both normals: $2 - 1 1 \times 1 - 2 3 = (- 1) (3) - (1) (- 2) (1) (1) - (2) (3) (2) (- 2) - (- 1) (1) = - 1 - 5 - 3 = - 153 .$ 2. Using the point $P$ from b) i): $L : r = 1 - 2 0 + λ 153 .$

c) 1. The normal of $Π_{3}$ is $(- 9, 3, - 2)$ and $(1, 5, 3) \cdot (- 9, 3, - 2) = - 9 + 15 - 6 = 0$ : the line is parallel to the plane, and since they do not intersect (part a), the line is not contained in it. 2. The distance between $L$ and $Π_{3}$ is the distance from any point of $L$ , for example $P (1, - 2, 0)$ , to the plane: $d = \frac{∣ - 9 \cdot 1 + 3 ( - 2 ) - 2 \cdot 0 - 32∣}{81 + 9 + 4} = \frac{47}{94} = \frac{94}{2} \approx 4.85.$

Answer: a) the system is inconsistent; b) ii) $r = 1 - 2 0 + λ 153$ ; c) $d = \frac{94}{2} \approx 4.85$ .

Exercise 135 HardIB · P2 TZ0 November 2008

Given the lines $r_{1} = 327 + m 2 - 1 2$ and $r_{2} = 142 + n 4 - 1 1$

a) Verify that these two lines intersect, and find the point of intersection, $A$ .

b) Find the Cartesian equation of the plane $π$ which contains these two lines.

c) Let $B$ be the point of intersection of the plane $π$ and the line $r = - 8 - 3 0 + λ 382$ . Find the coordinates of $B$ .

d) If $C$ is the midpoint of $A B$ , find the vector equation of the line which is perpendicular to the plane $π$ and passes through $C$ .

Show answer

Answer: a) $A (- 3, 5, 1)$ ; b) $x + 6 y + 2 z - 29 = 0$ ; c) $B (- 5, 5, 2)$ ; d) $r = - 4 5 3/2 + t 162$

Show working and answer

a) 1. We equate the coordinates: $3 + 2 m = 1 + 4 n, 2 - m = 4 - n, 7 + 2 m = 2 + n .$ 2. From the second, $n = m + 2$ . In the first: $3 + 2 m = 1 + 4 m + 8$ , so $m = - 3$ and $n = - 1$ . 3. We check in the third: $7 + 2 (- 3) = 1$ and $2 + (- 1) = 1$ . It holds, so the lines intersect. 4. With $m = - 3$ in $r_{1}$ : $A = (3 - 6, 2 + 3, 7 - 6) = (- 3, 5, 1)$ .

b) 1. A normal vector to the plane is the vector product of the direction vectors: $2 - 1 2 \times 4 - 1 1 = (- 1) \cdot 1 - 2 \cdot (- 1) 2 \cdot 4 - 2 \cdot 1 2 \cdot (- 1) - (- 1) \cdot 4 = 162 .$ 2. The plane is $x + 6 y + 2 z = d$ . With $A (- 3, 5, 1)$ : $d = - 3 + 30 + 2 = 29$ .

c) 1. The points of the line are $(- 8 + 3 λ, - 3 + 8 λ, 2 λ)$ . We substitute them into the plane: $(- 8 + 3 λ) + 6 (- 3 + 8 λ) + 2 (2 λ) = 29 \Rightarrow 55 λ - 26 = 29 \Rightarrow λ = 1.$ 2. $B = (- 5, 5, 2)$ .

d) 1. The midpoint of $A B$ is $C = (\frac{- 3 - 5}{2}, \frac{5 + 5}{2}, \frac{1 + 2}{2}) = (- 4, 5, \frac{3}{2})$ . 2. The direction vector is the normal $(1, 6, 2)$ : $r = - 4 5 3/2 + t 162 .$

Answer: a) $A (- 3, 5, 1)$ ; b) $x + 6 y + 2 z - 29 = 0$ ; c) $B (- 5, 5, 2)$ ; d) $r = - 4 5 3/2 + t 162$

Exercise 136 HardIB · P2 TZ2 May 2005

a) Let $π_{1}$ be the plane with equation $r = 211 + λ - 2 18 + μ 1 - 3 - 9$

Let $π_{2}$ be the plane with equation $r = 201 + s 121 + t 111$

(i) Verify that $λ = μ$ for the points common to $π_{1}$ and $π_{2}$ .

(ii) Hence, or otherwise, find a vector equation of the line of intersection of $π_{1}$ and $π_{2}$ .

b) The plane $π_{3}$ contains the line $\frac{2 - x}{3} = \frac{y}{- 4} = z + 1$ and is perpendicular to $3 i - 2 j + k$ .

Find the Cartesian equation of $π_{3}$ .

c) Find the intersection of $π_{1}$ , $π_{2}$ and $π_{3}$ .

Show answer

Answer: a) ii) $r = 211 + λ - 1 - 2 - 1$ ; b) $3 x - 2 y + z - 5 = 0$ ; c) the three planes intersect in the line from a) ii).

Show working and answer

a) i) 1. A common point satisfies $2 - 2 λ + μ = 2 + s + t, 1 + λ - 3 μ = 2 s + t, 1 + 8 λ - 9 μ = 1 + s + t .$ 2. The first equation gives $s + t = - 2 λ + μ$ and the third gives $s + t = 8 λ - 9 μ$ . Equating: $- 2 λ + μ = 8 λ - 9 μ$ , that is, $10 μ = 10 λ$ , so $λ = μ$ .

a) ii) 1. With $μ = λ$ the points of $π_{1}$ that also lie in $π_{2}$ are $2 - 2 λ + λ 1 + λ - 3 λ 1 + 8 λ - 9 λ = 2 - λ 1 - 2 λ 1 - λ .$ 2. Solving the system one checks that each $λ$ corresponds to values of $s$ and $t$ (with $t = - 1$ and $s = 1 - λ$ ), so all these points belong to $π_{2}$ . The line of intersection is $r = 211 + λ - 1 - 2 - 1 .$

b) 1. The line has parametric equations $x = 2 - 3 k$ , $y = - 4 k$ , $z = - 1 + k$ : it passes through $(2, 0, - 1)$ and its direction vector is $(- 3, - 4, 1)$ . 2. That the plane is perpendicular to the vector $3 i - 2 j + k$ means that this vector is normal to the plane: $3 x - 2 y + z = d$ . 3. With the point $(2, 0, - 1)$ : $d = 6 - 0 - 1 = 5$ . 4. We check that the plane contains the line: $(- 3, - 4, 1) \cdot (3, - 2, 1) = - 9 + 8 + 1 = 0$ .

c) 1. We substitute the line from a) ii) into $π_{3}$ : $3 (2 - λ) - 2 (1 - 2 λ) + (1 - λ) - 5 = 6 - 3 λ - 2 + 4 λ + 1 - λ - 5 = 0.$ 2. The result is $0 = 0$ for all $λ$ : the whole line lies in $π_{3}$ . The three planes intersect in that line.

Answer: a) ii) $r = 211 + λ - 1 - 2 - 1$ ; b) $3 x - 2 y + z - 5 = 0$ ; c) the three planes intersect in the line from a) ii).

Exercise 137 HardIB · P2 TZ2 May 2015

The equations of the lines $L_{1}$ and $L_{2}$ are

$L_{1} : r_{1} = 122 + λ - 1 12$ $L_{2} : r_{2} = 124 + μ 216$

a) Show that the lines $L_{1}$ and $L_{2}$ are skew.

b) Find the acute angle between the lines $L_{1}$ and $L_{2}$ .

c) (i) Find a vector perpendicular to both lines.

(ii) Hence, determine an equation of the line $L_{3}$ which is perpendicular to $L_{1}$ and $L_{2}$ and intersects both lines.

Show answer

Answer: b) $θ \approx 45. 5^{\circ}$ ; c) (i) $4 i + 10 j - 3 k$ ; (ii) $\frac{x - \frac{93}{125}}{4} = \frac{y - \frac{282}{125}}{10} = \frac{z - \frac{314}{125}}{- 3}$ . Part a) is proved.

Show working and answer

a) 1. The direction vectors $d_{1} = (- 1, 1, 2)$ and $d_{2} = (2, 1, 6)$ are not proportional, so the lines are not parallel. 2. Let's check whether they intersect: equate the coordinates, $1 - λ = 1 + 2 μ, 2 + λ = 2 + μ, 2 + 2 λ = 4 + 6 μ .$ 3. From the first equation, $λ = - 2 μ$ ; from the second, $λ = μ$ . So $μ = 0$ and $λ = 0$ . 4. In the third equation we get $2 = 4$ , which is false. There is no common point. 5. Lines that are neither parallel nor intersecting are skew.

b) 1. The acute angle between two lines satisfies $cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$ . 2. $d_{1} \cdot d_{2} = - 2 + 1 + 12 = 11$ , $∣ d_{1} ∣ = 6$ , $∣ d_{2} ∣ = 41$ . 3. $cos θ = \frac{11}{246}$ , so $θ \approx 45. 5^{\circ}$ .

c) (i) A vector perpendicular to both lines is the vector (cross) product $d_{1} \times d_{2} = - 1 12 \times 216 = 1 \cdot 6 - 2 \cdot 1 2 \cdot 2 - (- 1) \cdot 6 (- 1) \cdot 1 - 1 \cdot 2 = 410 - 3 .$

(ii) 1. $L_{3}$ has direction $n = (4, 10, - 3)$ . We look for $P$ on $L_{1}$ and $Q$ on $L_{2}$ such that $P Q$ is parallel to $n$ . 2. $P = (1 - λ, 2 + λ, 2 + 2 λ)$ and $Q = (1 + 2 μ, 2 + μ, 4 + 6 μ)$ , so $P Q = (2 μ + λ, μ - λ, 2 + 6 μ - 2 λ)$ . 3. We impose $P Q = c (4, 10, - 3)$ : $2 μ + λ = 4 c, μ - λ = 10 c, 2 + 6 μ - 2 λ = - 3 c .$ 4. Adding the first two, $3 μ = 14 c$ , that is $μ = \frac{14 c}{3}$ , and then $λ = μ - 10 c = - \frac{16 c}{3}$ . 5. In the third: $2 + 28 c + \frac{32 c}{3} = - 3 c$ , from which $\frac{125 c}{3} = - 2$ and $c = - \frac{6}{125}$ . Therefore $λ = \frac{32}{125}$ (and $μ = - \frac{28}{125}$ ). 6. $P = (\frac{93}{125}, \frac{282}{125}, \frac{314}{125})$ lies on $L_{1}$ and on $L_{3}$ . An equation of $L_{3}$ is $r = 93/125 282/125 314/125 + s 410 - 3, or \frac{x - \frac{93}{125}}{4} = \frac{y - \frac{282}{125}}{10} = \frac{z - \frac{314}{125}}{- 3} .$

Answer: b) $θ \approx 45. 5^{\circ}$ ; c) (i) $4 i + 10 j - 3 k$ ; (ii) $\frac{x - \frac{93}{125}}{4} = \frac{y - \frac{282}{125}}{10} = \frac{z - \frac{314}{125}}{- 3}$ . Part a) is proved.

Exercise 138 HardIB · P2 November 1996

(i) The two lines $L$ and $M$ and the plane $P$ are given by

$L : \frac{x - 3}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 1}$ $M : \frac{x - 6}{- 4} = \frac{y}{2} = z + 10$ $P : x + 3 y - 2 z = 4$ .

a) Calculate the coordinates of the point of intersection of the line $L$ and the plane $P$ .

b) Calculate the acute angle between the line $L$ and $P$ , giving your answer to the nearest tenth of a degree.

c) Show that the line $M$ is parallel to the plane $P$ .

d) Show that the lines $L$ and $M$ do not intersect.

e) Calculate the shortest distance between the lines $L$ and $M$ .

Show answer

Answer: a) $(4, 1, \frac{3}{2})$ ; b) $68. 9^{\circ}$ ; e) $d (L, M) = 110$ . Parts c) and d) are proved.

Show working and answer

a) 1. Parametric form of $L$ : $(3 + 2 t, - 1 + 4 t, 2 - t)$ . 2. Substituting into $P$ : $(3 + 2 t) + 3 (- 1 + 4 t) - 2 (2 - t) = 4 \Rightarrow 16 t - 4 = 4 \Rightarrow t = \frac{1}{2}$ . 3. Point: $(4, 1, \frac{3}{2})$ .

b) 1. Direction of $L$ : $d_{L} = (2, 4, - 1)$ ; normal of $P$ : $n = (1, 3, - 2)$ . 2. $sin θ = \frac{∣ d _{L} \cdot n ∣}{∣ d _{L} ∣∣ n ∣} = \frac{2 + 12 + 2}{21 14} = \frac{16}{294} \approx 0.933$ . 3. $θ \approx 68. 9^{\circ}$ .

c) 1. Direction of $M$ : $d_{M} = (- 4, 2, 1)$ (in $z + 10$ the denominator is $1$ ). 2. $d_{M} \cdot n = - 4 + 6 - 2 = 0$ : $M$ is parallel to $P$ or contained in it. 3. $M$ passes through $(6, 0, - 10)$ and $6 + 0 + 20 = 26 \neq = 4$ : that point is not on $P$ . Therefore $M$ is parallel to $P$ without being contained in it.

d) 1. $d_{L} = (2, 4, - 1)$ and $d_{M} = (- 4, 2, 1)$ are not proportional: the lines are not parallel. 2. If they intersected: $3 + 2 t = 6 - 4 s$ , $- 1 + 4 t = 2 s$ , $2 - t = - 10 + s$ . 3. From the second, $s = - \frac{1}{2} + 2 t$ ; in the first, $3 + 2 t = 6 - 4 (- \frac{1}{2} + 2 t) = 8 - 8 t$ , from which $10 t = 5$ , $t = \frac{1}{2}$ and $s = \frac{1}{2}$ . 4. In the third equation: $2 - \frac{1}{2} = \frac{3}{2}$ versus $- 10 + \frac{1}{2} = - \frac{19}{2}$ . They do not coincide. The lines do not intersect (they are skew).

e) 1. A vector perpendicular to both: $d_{L} \times d_{M} = 4 \cdot 1 - (- 1) \cdot 2 (- 1) (- 4) - 2 \cdot 1 2 \cdot 2 - 4 \cdot (- 4) = 6220$ , proportional to $(3, 1, 10)$ . 2. Points $A (3, - 1, 2) \in L$ and $B (6, 0, - 10) \in M$ : $A B = (3, 1, - 12)$ . 3. $d = \frac{∣ A B \cdot ( 3 , 1 , 10 ) ∣}{∣ ( 3 , 1 , 10 ) ∣} = \frac{∣9 + 1 - 120∣}{110} = \frac{110}{110} = 110 \approx 10.5$ .

Answer: a) $(4, 1, \frac{3}{2})$ ; b) $68. 9^{\circ}$ ; e) $d (L, M) = 110$ . Parts c) and d) are proved.

Exercise 139 HardIB · P2 May 1998

The equations of the planes $P_{1}$ and $P_{2}$ are given by $P_{1} : r \cdot (3 i - j + 2 k) = - 1 P_{2} : r \cdot (- 2 i + j - 5 k) = 4$ where $r = x i + y j + z k$ is the position vector of a point on the plane.

a) Let $L$ be the line of intersection of the two planes $P_{1}$ and $P_{2}$ .

(i) Show that $L$ is parallel to $3 i + 11 j + k$ .

(ii) Show that the point $A (0, - 1, - 1)$ lies on the line $L$ .

Using this, or otherwise, find the equation of $L$ .

The equation of a third plane $P_{3}$ is given by $P_{3} : r \cdot (- 4 i + j + k) = c$ .

b) Determine the value of $c$ for which the three planes $P_{1}$ , $P_{2}$ and $P_{3}$ intersect, and deduce whether this value of $c$ gives a point of intersection or a line of intersection.

c) For $c = 5$ ,

(i) Show that the plane $P_{3}$ is parallel to the line $L$ ;

(ii) Calculate the distance between the line $L$ and the plane $P_{3}$ .

Show answer

Answer: a) $L : r = 0 - 1 - 1 + t 3111$ ; b) $c = - 2$ , the three planes intersect in a line; c) $d (L, P_{3}) = \frac{7}{3 2} \approx 1.65$

Show working and answer

The Cartesian equations are $P_{1} : 3 x - y + 2 z = - 1$ , $P_{2} : - 2 x + y - 5 z = 4$ and $P_{3} : - 4 x + y + z = c$ , with normals $n_{1} = (3, - 1, 2)$ , $n_{2} = (- 2, 1, - 5)$ and $n_{3} = (- 4, 1, 1)$ .

a) i) 1. The line $L$ lies in both planes, so its direction is perpendicular to $n_{1}$ and to $n_{2}$ . A vector with that property is their cross product: $n_{1} \times n_{2} = (- 1) (- 5) - 2 \cdot 1 2 (- 2) - 3 (- 5) 3 \cdot 1 - (- 1) (- 2) = 3111 .$ 2. Hence $L$ is parallel to $3 i + 11 j + k$ .

(ii) 1. Substituting $A (0, - 1, - 1)$ into $P_{1}$ : $3 \cdot 0 - (- 1) + 2 (- 1) = 1 - 2 = - 1$ . It holds. 2. In $P_{2}$ : $- 2 \cdot 0 + (- 1) - 5 (- 1) = - 1 + 5 = 4$ . It holds. 3. As $A$ lies in both planes, $A \in L$ , and the line passes through $A$ with direction $(3, 11, 1)$ : $L : r = 0 - 1 - 1 + t 3111, x = 3 t, y = - 1 + 11 t, z = - 1 + t .$

b) 1. The three planes have a common point if $P_{3}$ has one with $L$ (which is the intersection of $P_{1}$ and $P_{2}$ ). Substituting the points of $L$ into $P_{3}$ : $- 4 (3 t) + (- 1 + 11 t) + (- 1 + t) = - 12 t + 11 t + t - 2 = - 2.$ 2. The result does not depend on $t$ . Therefore $- 2 = c$ is the only condition: if $c = - 2$ , all points of $L$ lie in $P_{3}$ ; if $c \neq = - 2$ , no point of $L$ lies in $P_{3}$ . 3. For the three planes to intersect, $c = - 2$ , and then their intersection is the whole line $L$ : it is a line of intersection.

c) i) 1. With $c = 5$ , the direction vector of $L$ satisfies $(3, 11, 1) \cdot (- 4, 1, 1) = - 12 + 11 + 1 = 0$ , so $L$ is perpendicular to $n_{3}$ , that is, parallel to the plane $P_{3}$ or contained in it. 2. As $A$ is not in $P_{3}$ (it gives $- 2 \neq = 5$ ), $L$ is not contained in $P_{3}$ . Therefore $L$ is parallel to $P_{3}$ .

(ii) 1. The distance from the line to the plane is that of any of its points to the plane; we use $A (0, - 1, - 1)$ and $P_{3} : - 4 x + y + z - 5 = 0$ : $d = \frac{∣ - 4 \cdot 0 + ( - 1 ) + ( - 1 ) - 5∣}{16 + 1 + 1} = \frac{7}{18} = \frac{7}{3 2} \approx 1.65.$

Answer: a) $L : r = 0 - 1 - 1 + t 3111$ ; b) $c = - 2$ , the three planes intersect in a line; c) $d (L, P_{3}) = \frac{7}{3 2} \approx 1.65$

Exercise 140 HardIB · P2 May 1997

a) The line $L_{1}$ is parallel to the vector $v = 3 i + j + 3 k$ and passes through the point $(2, 3, 7)$ .

Calculate the vector equation of the line.

b) The equation of the plane, $E$ , is given by $2 x + 3 y - 4 z + 21 = 0$ .

Calculate the point of intersection of the line $L_{1}$ and the plane $E$ .

c) Calculate the equation of the plane that passes through the point $(1, 2, 3)$ and is parallel to the plane $E$ .

d) The parametric equations of another line $L_{2}$ are $x = t, y = t, z = - t, - \infty < t < \infty$ .

Show that

(i) $L_{1}$ is not parallel to $L_{2}$ ;

(ii) $L_{1}$ does not intersect $L_{2}$ .

e) Let $O$ be the origin and $P$ the point $(- 1, 2, 4)$ .

(i) Calculate a vector $w$ parallel to the line $L_{2}$ .

(ii) Calculate the vector $P O$ .

(iii) Calculate the shortest distance between the lines $L_{1}$ and $L_{2}$ using the formula

$d = \frac{P O \cdot ( v \times w )}{∣ v \times w ∣} .$

Show answer

Answer: a) $r = 237 + λ 313$ ; b) $(8, 5, 13)$ ; c) $2 x + 3 y - 4 z + 4 = 0$ ; e) i) $w = (1, 1, - 1)$ , ii) $P O = (1, - 2, - 4)$ , iii) $d = \frac{12}{14} \approx 3.21$

Show working and answer

a) 1. The line passes through $(2, 3, 7)$ with direction $v = (3, 1, 3)$ : $L_{1} : r = 237 + λ 313 .$

b) 1. A general point on $L_{1}$ is $(2 + 3 λ, 3 + λ, 7 + 3 λ)$ . We substitute it into $E$ : $2 (2 + 3 λ) + 3 (3 + λ) - 4 (7 + 3 λ) + 21 = 4 + 6 λ + 9 + 3 λ - 28 - 12 λ + 21 = 6 - 3 λ = 0.$ 2. Hence $λ = 2$ and the point of intersection is $(8, 5, 13)$ .

c) 1. A plane parallel to $E$ has the same normal: $2 x + 3 y - 4 z + D = 0$ . 2. It passes through $(1, 2, 3)$ : $2 + 6 - 12 + D = 0 \Rightarrow D = 4$ . 3. The plane is $2 x + 3 y - 4 z + 4 = 0$ .

d) i) 1. The directions are $(3, 1, 3)$ and $(1, 1, - 1)$ . If they were proportional, $(3, 1, 3) = k (1, 1, - 1)$ ; the first component would require $k = 3$ and the second $k = 1$ , which is impossible. Therefore $L_{1}$ is not parallel to $L_{2}$ .

(ii) 1. A common point would satisfy $2 + 3 λ = t$ , $3 + λ = t$ , $7 + 3 λ = - t$ . 2. From the first two: $2 + 3 λ = 3 + λ \Rightarrow λ = \frac{1}{2}$ and $t = \frac{7}{2}$ . 3. In the third: $7 + \frac{3}{2} = \frac{17}{2}$ and $- t = - \frac{7}{2}$ . These do not match, so there is no solution: $L_{1}$ does not intersect $L_{2}$ .

e) i) 1. The direction of $L_{2}$ is $w = 11 - 1$ .

(ii) 1. $P O = O - P = 1 - 2 - 4$ .

(iii) 1. We calculate $v \times w = 313 \times 11 - 1 = 1 \cdot (- 1) - 3 \cdot 1 3 \cdot 1 - 3 \cdot (- 1) 3 \cdot 1 - 1 \cdot 1 = - 4 62, ∣ v \times w ∣ = 56 = 214 .$ 2. The scalar product is $P O \cdot (v \times w) = 1 (- 4) + (- 2) (6) + (- 4) (2) = - 24$ . 3. Then $d = \frac{∣ - 24∣}{2 14} = \frac{12}{14} = \frac{6 14}{7} \approx 3.21.$ (Here $P$ lies on $L_{1}$ , with $λ = - 1$ , and $O$ lies on $L_{2}$ , with $t = 0$ .)

Answer: a) $r = 237 + λ 313$ ; b) $(8, 5, 13)$ ; c) $2 x + 3 y - 4 z + 4 = 0$ ; e) i) $w = (1, 1, - 1)$ , ii) $P O = (1, - 2, - 4)$ , iii) $d = \frac{12}{14} \approx 3.21$

Exercise 141 HardIB · P2 TZ2 May 2012

a) Find the values of $k$ for which the following system of equations has no solution, and the value of $k$ for which the system has infinitely many solutions.

$x - 3 y + z x + 5 y - 2 z 16 y - 6 z = 3 = 1 = k$

b) Given that the system of equations can be solved, find the solutions in the form of a vector equation of a line, $r = a + λ b$ , where the components of $b$ are integers.

c) The plane $π$ is parallel to both the line in part (b) and the line $\frac{x - 4}{3} = \frac{y - 6}{- 2} = \frac{z - 2}{0}$ .

Given that the point $(1, 2, 0)$ belongs to $π$ , verify that the Cartesian equation of $π$ is $16 x + 24 y - 11 z = 64$ .

d) The $z$ -axis meets the plane $π$ at the point $P$ . Find the coordinates of $P$ .

e) Find the angle between the line $\frac{x - 2}{3} = \frac{y + 5}{4} = \frac{z}{2}$ and the plane $π$ .

Show answer

Answer: a) no solution if $k \neq = - 4$ ; infinitely many solutions if $k = - 4$ . b) $r = \frac{9}{4} - \frac{1}{4} 0 + λ 138$ . c) $16 x + 24 y - 11 z = 64$ . d) $P (0, 0, - \frac{64}{11})$ . e) $θ \approx 47. 2^{\circ}$ .

Show working and answer

a) 1. Subtracting the first two equations: $(x + 5 y - 2 z) - (x - 3 y + z) = 1 - 3$ , that is, $8 y - 3 z = - 2$ . 2. Multiplying by $2$ : $16 y - 6 z = - 4$ . 3. The third equation says $16 y - 6 z = k$ . If $k \neq = - 4$ it contradicts the previous one and the system has no solution. If $k = - 4$ the third equation follows from the other two, and two equations remain in three unknowns: infinitely many solutions.

b) 1. With $k = - 4$ the system is equivalent to $x - 3 y + z = 3$ and $8 y - 3 z = - 2$ . 2. Take $z = t$ : $y = \frac{3 t - 2}{8}$ . 3. $x = 3 + 3 y - z = 3 + \frac{9 t - 6}{8} - t = \frac{18 + t}{8}$ . 4. Therefore $(x, y, z) = (\frac{9}{4}, - \frac{1}{4}, 0) + \frac{t}{8} (1, 3, 8)$ . Taking $λ = \frac{t}{8}$ : $r = \frac{9}{4} - \frac{1}{4} 0 + λ 138 .$

c) 1. The directions of the two lines are $b = 138$ and $c = 3 - 2 0$ . The normal of the plane is perpendicular to both: $b \times c = 3 \cdot 0 - 8 \cdot (- 2) 8 \cdot 3 - 1 \cdot 0 1 \cdot (- 2) - 3 \cdot 3 = 1624 - 11 .$ 2. Therefore $π : 16 x + 24 y - 11 z = d$ . With $(1, 2, 0)$ : $d = 16 + 48 = 64$ . 3. $π : 16 x + 24 y - 11 z = 64$ .

d) 1. The points of the $z$ -axis are $(0, 0, z)$ . Substituting: $- 11 z = 64$ , so $z = - \frac{64}{11}$ . 2. $P = (0, 0, - \frac{64}{11})$ .

e) 1. The direction of the line is $d = 342$ and the normal of the plane is $n = 1624 - 11$ . 2. $d \cdot n = 48 + 96 - 22 = 122$ , $∣ d ∣ = 29$ and $∣ n ∣ = 256 + 576 + 121 = 953$ . 3. $sin θ = \frac{122}{29 953} = \frac{122}{27637} \approx 0.734$ , so $θ \approx 47. 2^{\circ}$ .

Exercise 142 HardIB · P2 TZ1 May 2011

The equations of three planes are given by

$a x + 2 y + z = 3$

$- x + (a + 1) y + 3 z = 1, a \in R$

$- 2 x + y + (a + 2) z = k$

a) Given that $a = 0$ , show that the three planes intersect at a point.

b) Find the value of $a$ such that the three planes do not intersect at a point.

c) Given that the three planes do not intersect at a point, find the value of $k$ such that the planes intersect in a line, and find an equation of this line in the form $x y z = x_{0} y_{0} z_{0} + λ l m n$ .

Show answer

Answer: a) the determinant is $- 7 \neq = 0$ , so they intersect at one point; b) $a = 1$ ; c) $k = - 1$ and $x y z = 110 + λ 1 - 1 1$ .

Show working and answer

a) 1. The three planes intersect at a point when the determinant of the coefficient matrix is not zero. With $a = 0$ : $0 - 1 - 2 211132 = 0 \cdot (2 - 3) - 2 \cdot (- 2 + 6) + 1 \cdot (- 1 + 2) = - 8 + 1 = - 7.$ 2. As $- 7 \neq = 0$ , the system is consistent and determined: the three planes intersect at a unique point.

b) 1. We calculate the determinant for a general value of $a$ (using the GDC or by hand): $a - 1 - 2 2 a + 1 1 13 a + 2 = (a - 1) (a^{2} + 4 a + 7) .$ 2. The factor $a^{2} + 4 a + 7$ never vanishes (its discriminant is $16 - 28 < 0$ ). The determinant is zero only if $a = 1$ , which is the required value.

c) 1. With $a = 1$ the system is: $x + 2 y + z = 3, - x + 2 y + 3 z = 1, - 2 x + y + 3 z = k .$ 2. Adding the first two: $4 y + 4 z = 4$ , that is $y + z = 1$ . 3. Subtracting the third from the second: $x + y = 1 - k$ . 4. Substituting $z = 1 - y$ into the first: $x + 2 y + 1 - y = 3$ , that is, $x + y = 2$ . 5. For the system to be consistent we need $1 - k = 2$ , that is, $k = - 1$ . Then the system reduces to two independent equations and the planes intersect in a line. 6. Taking $z = λ$ : $y = 1 - λ$ and $x = 2 - y = 1 + λ$ . $x y z = 110 + λ 1 - 1 1 .$

Answer: a) the determinant is $- 7 \neq = 0$ , so they intersect at one point; b) $a = 1$ ; c) $k = - 1$ and $x y z = 110 + λ 1 - 1 1$ .

Contextual problems

Exercise 143 MediumIB · P1 TZ0 November 2012

Two boats, $A$ and $B$ , move so that at time $t$ hours their position vectors, in kilometres, are $r_{A} = (9 t) i + (3 - 6 t) j$ and $r_{B} = (7 - 4 t) i + (7 t - 6) j$ .

a) Find the coordinates of the common point of the paths of the two boats.

b) Show that the boats do not collide.

Show answer

Answer: a) $(3, 1)$ ; b) $A$ passes through $(3, 1)$ at $t = \frac{1}{3}$ h and $B$ at $t = 1$ h; these are different times, so the boats do not collide.

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a) 1. The path of $A$ has equations $x = 9 t$ , $y = 3 - 6 t$ . Eliminating the time ( $t = \frac{x}{9}$ ): $y = 3 - \frac{2 x}{3}$ . 2. The path of $B$ has $x = 7 - 4 t$ , $y = 7 t - 6$ . With $t = \frac{7 - x}{4}$ : $y = \frac{7 ( 7 - x )}{4} - 6 = \frac{25 - 7 x}{4}$ . 3. Equating: $3 - \frac{2 x}{3} = \frac{25 - 7 x}{4}$ . Multiplying by $12$ : $36 - 8 x = 75 - 21 x$ , so $13 x = 39$ and $x = 3$ . 4. Then $y = 3 - 2 = 1$ . The common point of the paths is $(3, 1)$ .

b) 1. Boat $A$ is at $(3, 1)$ when $9 t = 3$ , that is $t = \frac{1}{3}$ h (and with this value $3 - 6 t = 1$ , which is correct). 2. Boat $B$ is at $(3, 1)$ when $7 - 4 t = 3$ , that is $t = 1$ h (and with this value $7 t - 6 = 1$ , which is correct). 3. They pass the point of intersection at different instants ( $\frac{1}{3}$ h and $1$ h), so they are never at the same place at the same time: the boats do not collide.

Answer: a) $(3, 1)$ ; b) $A$ passes through $(3, 1)$ at $t = \frac{1}{3}$ h and $B$ at $t = 1$ h; these are different times, so the boats do not collide.

Exercise 144 MediumIB · P1 TZ0 November 2022

Consider a circle with diameter $A B$ , where $A$ has coordinates $(1, 4, 0)$ and $B$ has coordinates $(- 3, 2, - 4)$ .

a) Find:

(i) The coordinates of the centre of the circle

(ii) The radius of the circle

The circle is the base of a right circular cone whose vertex $V$ has coordinates $(- 1, - 1, 0)$ .

b) Find the exact volume of the cone.

Show answer

Answer: a) i) centre $(- 1, 3, - 2)$ ; ii) radius $3$ ; b) $V = 65 π$ (cubic units).

Show working and answer

a) (i) 1. The centre is the midpoint of $[A B]$ : $M = (\frac{1 + ( - 3 )}{2}, \frac{4 + 2}{2}, \frac{0 + ( - 4 )}{2}) = (- 1, 3, - 2) .$

(ii) 1. The diameter has length $∣ A B ∣ = - 4 - 2 - 4 = 16 + 4 + 16 = 6$ . 2. The radius is half of this: $r = 3$ .

b) 1. As the cone is right, the height is the distance from the vertex $V$ to the centre $M$ of the base: $M V = 0 - 4 2$ , so $h = 0 + 16 + 4 = 20 = 25$ . 2. The volume of the cone is $\frac{1}{3} π r^{2} h$ : $V = \frac{1}{3} π \cdot 3^{2} \cdot 25 = 65 π \approx 42.1.$

Answer: a) i) centre $(- 1, 3, - 2)$ ; ii) radius $3$ ; b) $V = 65 π$ (cubic units).

Exercise 145 HardIB · P2 TZ1 May 2012

The coordinates of the points $A$ , $B$ and $C$ are $(5, - 2, 5)$ , $(5, 4, - 1)$ and $(- 1, - 2, - 1)$ , respectively.

a) Show that $A B = A C$ and that $B A C = 6 0^{\circ}$ .

b) Find the Cartesian equation of $Π$ , the plane through $A$ , $B$ and $C$ .

c) (i) Find the Cartesian equation of $Π_{1}$ , the plane perpendicular to $(A B)$ passing through the midpoint of $[A B]$ .

(ii) Find the Cartesian equation of $Π_{2}$ , the plane perpendicular to $(A C)$ passing through the midpoint of $[A C]$ .

d) Find the vector equation of $L$ , the line of intersection of $Π_{1}$ and $Π_{2}$ , and show that it is perpendicular to $Π$ .

A methane molecule consists of one carbon atom with four hydrogen atoms symmetrically arranged around it in three dimensions.

The positions of the centres of three of the hydrogen atoms are $A$ , $B$ and $C$ , as given in the diagram. The position of the centre of the fourth hydrogen atom is $D$ .

e) Using the fact that $A B = A D$ , show that the coordinates of one of the possible positions of the fourth hydrogen atom are $(- 1, 4, 5)$ .

f) Given that $D (- 1, 4, 5)$ , show that the coordinates of $G$ , the position of the centre of the carbon atom, are $(2, 1, 2)$ .

Using the above, calculate $D G A$ , the bond angle of the carbon.

Show answer

Answer: b) $Π : - x + y + z + 2 = 0$ ; c) i) $Π_{1} : y - z + 1 = 0$ ; ii) $Π_{2} : - x - z + 4 = 0$ ; d) $L : r = 4 - 1 0 + t - 1 11$ ; f) $G = (2, 1, 2)$ and $D G A \approx 109. 5^{\circ}$ . Parts a), d), e) and f) are proved.

Show working and answer

a) 1. $A B = 06 - 6$ and $A C = - 6 0 - 6$ . 2. $A B = 0 + 36 + 36 = 62$ and $A C = 36 + 0 + 36 = 62$ . Therefore $A B = A C$ . 3. Scalar product: $A B \cdot A C = 0 + 0 + 36 = 36$ . 4. $cos B A C = \frac{A B \cdot A C}{A B \cdot A C} = \frac{36}{72} = \frac{1}{2}$ , so $B A C = 6 0^{\circ}$ .

b) 1. A normal vector is $A B \times A C = 6 (- 6) - (- 6) (0) (- 6) (- 6) - 0 \cdot (- 6) 0 \cdot 0 - 6 (- 6) = - 36 3636 = 36 - 1 11$ . 2. The plane is $- x + y + z = d$ . With $A$ : $d = - 5 - 2 + 5 = - 2$ . 3. $Π : - x + y + z + 2 = 0$ . Check: $B$ : $- 5 + 4 - 1 = - 2$ and $C$ : $1 - 2 - 1 = - 2$ .

c) (i) 1. The plane $Π_{1}$ has normal $A B$ , parallel to $(0, 1, - 1)$ : $y - z = d$ . 2. The midpoint of $[A B]$ is $M_{1} = (5, 1, 2)$ , so $d = 1 - 2 = - 1$ . 3. $Π_{1} : y - z + 1 = 0$ .

(ii) 1. The normal of $Π_{2}$ is $A C$ , parallel to $(1, 0, 1)$ : $x + z = d$ . 2. The midpoint of $[A C]$ is $M_{2} = (2, - 2, 2)$ , so $d = 2 + 2 = 4$ . 3. $Π_{2} : - x - z + 4 = 0$ (that is, $x + z = 4$ ).

d) 1. The points of $L$ satisfy $y - z = - 1$ and $x + z = 4$ . Taking $z = t$ as the parameter: $y = t - 1$ , $x = 4 - t$ . 2. Vector equation: $L : r = 4 - 1 0 + t - 1 11 .$ 3. The direction vector $(- 1, 1, 1)$ is exactly the normal of $Π$ calculated in b). A line whose direction vector is parallel to the normal of a plane is perpendicular to that plane.

e) 1. $Π_{1}$ and $Π_{2}$ are the planes formed by the points equidistant from $A$ and $B$ , and from $A$ and $C$ , respectively, so $L$ consists of the points equidistant from $A$ , $B$ and $C$ . As $A B = A C = A D$ and the four atoms are placed symmetrically, $A B = B C = A C = A D = B D = C D$ , and $D$ is equidistant from $A$ , $B$ and $C$ , that is, $D$ lies on $L$ . 2. A point of $L$ is $D = (4 - t, t - 1, t)$ . Then $A D = - 1 - t t + 1 t - 5$ . 3. Impose $A D^{2} = A B^{2} = 72$ : $2 (t + 1)^{2} + (t - 5)^{2} = 72 \Rightarrow 3 t^{2} - 6 t - 45 = 0 \Rightarrow t^{2} - 2 t - 15 = 0.$ 4. The solutions are $t = 5$ and $t = - 3$ . With $t = 5$ we obtain $D = (- 1, 4, 5)$ (and with $t = - 3$ , $D = (7, - 4, - 3)$ , the symmetric position on the other side of the plane $Π$ ). Therefore $(- 1, 4, 5)$ is one of the possible positions.

f) 1. In a symmetric molecule the centre $G$ of the carbon is equidistant from the four hydrogen atoms. Being equidistant from $A$ , $B$ and $C$ means lying on $L$ : $G = (4 - s, s - 1, s)$ . 2. Impose $G A = G D$ : $G A^{2} = (- 1 - s)^{2} + (s + 1)^{2} + (s - 5)^{2} = 3 s^{2} - 6 s + 27, G D^{2} = (5 - s)^{2} + (s - 5)^{2} + (s - 5)^{2} = 3 s^{2} - 30 s + 75.$ 3. Equating: $- 6 s + 27 = - 30 s + 75$ , so $24 s = 48$ and $s = 2$ . Then $G = (2, 1, 2)$ . 4. Check: $G A = (3, - 3, 3)$ , $GB = (3, 3, - 3)$ , $GC = (- 3, - 3, - 3)$ , $G D = (- 3, 3, 3)$ ; all four have modulus $33$ . 5. Bond angle: $G D \cdot G A = - 9 - 9 + 9 = - 9$ and $∣ G D ∣∣ G A ∣ = 27$ , so $cos D G A = \frac{- 9}{27} = - \frac{1}{3} \Rightarrow D G A = arccos (- \frac{1}{3}) \approx 109. 5^{\circ} .$

Exercise 146 HardIB · P2 TZ2 May 2022

Two aeroplanes ( $A$ and $B$ ) have position vectors relative to the origin $O$ , respectively

$r_{A} = 19 - 1 1 + t - 6 24 r_{B} = 1012 + t 42 - 2$

where $t$ represents time in minutes and $0 \leq t \leq 2.5$ .

The components of each column vector represent the displacement east of $O$ , the displacement north of $O$ and the height above sea level, all measured in kilometres.

a) Find the bearing, expressed to three figures, on which aeroplane $B$ is travelling.

b) Show that aeroplane $A$ travels at a greater speed than aeroplane $B$ .

c) Find the acute angle between the flight paths of the two aeroplanes.

Give your answer in degrees.

The flight paths of the two aeroplanes intersect at the point $P$ .

d) (i) Find the coordinates of $P$ .

(ii) Determine the time that elapses between the first aeroplane arriving at $P$ and the second aeroplane arriving at $P$ .

Let $D (t)$ be the distance between aeroplane $A$ and aeroplane $B$ for $0 \leq t \leq 2.5$ .

e) Find the minimum value of $D (t)$ .

Show answer

Answer: a) $06 3^{\circ}$ ; b) $v_{A} \approx 7.48$ km/min $> v_{B} \approx 4.90$ km/min; c) $40. 2^{\circ}$ ; d) i) $P (7, 3, 9)$ ; ii) $0.5$ minutes; e) $D_{m i n} \approx 1.01$ km (for $t \approx 1.81$ min).

Show working and answer

a) 1. The velocity vector of $B$ is $(4, 2, - 2)$ : $4$ km/min to the east and $2$ km/min to the north (the vertical component does not affect the bearing). 2. The bearing is measured clockwise from north. The angle from north satisfies $tan θ = \frac{east}{north} = \frac{4}{2} = 2$ . 3. $θ = arctan 2 \approx 63. 4^{\circ}$ . To three figures, the bearing is $06 3^{\circ}$ .

b) 1. The speed is the magnitude of the velocity vector. For $A$ : $∣ (- 6, 2, 4) ∣ = 36 + 4 + 16 = 56 = 214 \approx 7.48$ km/min. 2. For $B$ : $∣ (4, 2, - 2) ∣ = 16 + 4 + 4 = 24 = 26 \approx 4.90$ km/min. 3. As $7.48 > 4.90$ , aeroplane $A$ travels faster than aeroplane $B$ .

c) 1. The acute angle between the paths satisfies $cos θ = \frac{∣ d _{A} \cdot d _{B} ∣}{∣ d _{A} ∣∣ d _{B} ∣}$ . 2. $d_{A} \cdot d_{B} = - 24 + 4 - 8 = - 28$ , and $∣ d_{A} ∣∣ d_{B} ∣ = 5624 = 1344$ . 3. $cos θ = \frac{28}{1344} \approx 0.7638$ , so $θ \approx 40. 2^{\circ}$ .

d) (i) 1. We equate the positions at different times ( $t$ for $A$ and $s$ for $B$ ), because the paths can cross without the aeroplanes being at the same place at the same time: $19 - 6 t = 1 + 4 s, - 1 + 2 t = 2 s, 1 + 4 t = 12 - 2 s .$ 2. From the second, $s = t - \frac{1}{2}$ . Substituting into the first: $19 - 6 t = 1 + 4 t - 2$ , so $10 t = 20$ , $t = 2$ and $s = \frac{3}{2}$ . 3. Check in the third: $1 + 8 = 9$ and $12 - 3 = 9$ . Correct. 4. $P = (19 - 12, - 1 + 4, 1 + 8) = (7, 3, 9)$ .

(ii) 1. Aeroplane $B$ reaches $P$ at $s = 1.5$ min and aeroplane $A$ reaches it at $t = 2$ min. 2. Between the two arrivals, $2 - 1.5 = 0.5$ minutes elapse.

e) 1. The vector between the aeroplanes at the same time $t$ is $r_{A} - r_{B} = 18 - 10 t - 1 6 t - 11$ . 2. $D (t)^{2} = (18 - 10 t)^{2} + 1 + (6 t - 11)^{2} = 136 t^{2} - 492 t + 446$ . 3. $D$ is minimal where $D^{2}$ is (a parabola with its minimum at the vertex): $t = \frac{492}{272} = \frac{123}{68} \approx 1.81$ min, which lies within $0 \leq t \leq 2.5$ . It can also be found using the GDC by minimising $D (t)$ . 4. $D^{2} (\frac{123}{68}) = 446 - \frac{49 2 ^{2}}{4 \cdot 136} = \frac{35}{34}$ , so $D_{m i n} = \frac{35}{34} \approx 1.01$ km.

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