IB Analysis and Approaches HL · Geometry in space · Section 5 of 7

Angles and distances in space

3 worked examples · By Álvaro López

Key idea

The scalar (dot) product measures angles between lines and planes, and the normal vector, or the vector product, measures distances. Each calculation starts from a direction vector or a normal vector and ends in a short formula. In the IB exam these appear on their own in Paper 1 and combined in the longer Paper 2 problems.

Angle between two lines

Two lines that intersect form four angles, equal in pairs: one acute (or right) and one obtuse. The angle between the lines is always the acute one, that is, a value between $0^{\circ}$ and $9 0^{\circ}$ . If the lines are parallel, the angle is $0^{\circ}$ .

Two lines that pass each other without meeting (called skew lines) do not form an angle in the strict sense, but they are assigned the angle of their directions: the angle formed by two intersecting lines that have the same direction vectors.

A direction vector sets the direction of travel, and its opposite describes the same line. That is why, when applying the dot product, the absolute value is taken, which discards the obtuse angle.

This follows directly from the formula for the angle between vectors in the scalar (dot) product section, with the absolute value added. If you leave out the absolute value and the dot product is negative, you get the obtuse angle: $18 0^{\circ} - θ$ .

Angle between two planes and between a line and a plane

A plane is oriented by its normal vector $n$ , the one perpendicular to it (in $a x + b y + cz = d$ it is $n = (a, b, c)$ ; if you do not remember this, review the lines and planes section). Two intersecting planes form an angle equal to that of their normal vectors, because rotating a plane rotates its normal vector by the same angle.

For a line and a plane the situation changes. The angle between the line and the plane is the one formed by the line and its projection onto the plane (the line's "shadow"). If $φ$ is the angle between the line and the plane, the angle between the line and the normal is $9 0^{\circ} - φ$ , and since $cos (9 0^{\circ} - φ) = sin φ$ , the sine appears.

Special cases: if $d \cdot n = 0$ , the line is parallel to the plane (or lies in it) and $φ = 0^{\circ}$ ; if $d$ is proportional to $n$ , the line is perpendicular to the plane and $φ = 9 0^{\circ}$ .

Distance from a point to a line

The distance from a point $P$ to a line $ℓ$ is the shortest distance from $P$ to a point on $ℓ$ . It is reached at the point $Q$ on the line such that $P Q$ is perpendicular to it. There are two methods, and it is worth mastering both.

This method also gives the foot $Q$ , which is often asked for separately (for example, to find the reflection of $P$ in the line).

The reason: $∣ A P \times d ∣$ is the area of the parallelogram with sides $A P$ and $d$ . Its base has length $∣ d ∣$ and its height is exactly the distance sought, so dividing the area by the base gives the height.

For two parallel lines, the distance between them is the distance from any point of one to the other. For a line parallel to a plane, it is the distance from any point of the line to the plane.

Distance from a point to a plane

The distance from $P$ to a plane $π$ is the length of the segment perpendicular to $π$ that starts at $P$ . If $Q$ is its foot on the plane, $d = ∣ P Q ∣$ .

Plane pi with a point F on the plane and a point P outside it; the distance d is the length of PQ, perpendicular to the plane, and is the projection of FP onto the normal n

To obtain a formula, take any point $F$ on the plane. The distance $d$ is the length of the projection of $F P$ onto the direction of the normal $n$ (in the diagram, the dashed segment $F P$ and the perpendicular $P Q$ ). That length is $\frac{∣ F P \cdot n ∣}{∣ n ∣}$ , and in the numerator, with $F$ on the plane, it becomes $p \cdot n - f \cdot n = p \cdot n - d_{0}$ .

Notice that the plane is written with the constant term on the right-hand side, so the numerator subtracts $d$ . If the equation comes as $a x + b y + cz + k = 0$ , the numerator is $∣ a x_{0} + b y_{0} + c z_{0} + k ∣$ .

The same idea gives the distance between two parallel planes (take a point on one and substitute it into the equation of the other).

Distance between two skew lines

If two lines $ℓ_{1}$ and $ℓ_{2}$ are neither parallel nor intersecting, the distance between them is measured along the common perpendicular: the only segment that joins the two lines and is perpendicular to both. Its direction is that of a vector perpendicular to $d_{1}$ and $d_{2}$ , that is, $d_{1} \times d_{2}$ .

It is the projection of the vector $A_{1} A_{2}$ (which joins the two lines) onto the direction of the common perpendicular. If $d_{1} \times d_{2} = 0$ , the lines are parallel and the formula does not work: use the distance from a point to a line. If the numerator is $0$ , the lines intersect.

It is the same calculation as the formula, but step by step, and it serves as a check.

Worked examples

Example 1 Basic

Angles between lines and planes

Consider the lines $ℓ_{1} : r = 102 + λ 122$ and $ℓ_{2} : r = 3 - 1 0 + μ 2 - 1 2$ , and the planes $π_{1} : 2 x + 3 y + 6 z = 12$ and $π_{2} : x - 2 y + 2 z = 4$ .

a) Find the angle between $ℓ_{1}$ and $ℓ_{2}$ .

b) Find the angle between $ℓ_{1}$ and $π_{1}$ .

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

Give each angle in degrees, to one decimal place.

Solution

a) The direction vectors are $d_{1} = (1, 2, 2)$ and $d_{2} = (2, - 1, 2)$ , with $∣ d_{1} ∣ = 3$ and $∣ d_{2} ∣ = 3$ . The dot product is $d_{1} \cdot d_{2} = 2 - 2 + 4 = 4$ :

$cos θ = \frac{∣ 4 ∣}{3 \cdot 3} = \frac{4}{9} ⟹ θ \approx 63. 6^{\circ}$

b) With $n_{1} = (2, 3, 6)$ , $∣ n_{1} ∣ = 4 + 9 + 36 = 7$ , and $d_{1} \cdot n_{1} = 2 + 6 + 12 = 20$ . Since this is a line and a plane, the sine is used:

$sin φ = \frac{20}{3 \cdot 7} = \frac{20}{21} ⟹ φ \approx 72. 2^{\circ}$

c) With $n_{2} = (1, - 2, 2)$ , $∣ n_{2} ∣ = 3$ , and $n_{1} \cdot n_{2} = 2 - 6 + 12 = 8$ :

$cos θ = \frac{8}{7 \cdot 3} = \frac{8}{21} ⟹ θ \approx 67. 6^{\circ}$

Example 2 Medium

Distances from a point: to a line and to a plane

Consider the point $P (4, 3, - 1)$ , the line $ℓ : r = 120 + λ 21 - 2$ and the plane $π : x + 2 y - 2 z = 3$ .

a) Find the foot $Q$ of the perpendicular from $P$ to the line $ℓ$ .

b) Calculate the distance from $P$ to $ℓ$ .

c) Calculate the distance from $P$ to the plane $π$ .

d) Find the point on $π$ closest to $P$ .

Solution

a) A general point on $ℓ$ is $Q (1 + 2 λ, 2 + λ, - 2 λ)$ . Then

$P Q = 2 λ - 3 λ - 1 1 - 2 λ$

and we impose $P Q \cdot d = 0$ with $d = (2, 1, - 2)$ :

$2 (2 λ - 3) + (λ - 1) - 2 (1 - 2 λ) = 9 λ - 9 = 0 ⟹ λ = 1$

Therefore $Q (3, 3, - 2)$ .

b) $P Q = (- 1, 0, - 1)$ , so $dist (P, ℓ) = 1 + 0 + 1 = 2$ .

Check with the vector product: with $A (1, 2, 0)$ , $A P = (3, 1, - 1)$ and $A P \times d = (- 1, 4, 1)$ , of magnitude $18 = 32$ . Dividing by $∣ d ∣ = 3$ gives $2$ .

c) With $n = (1, 2, - 2)$ , $∣ n ∣ = 3$ :

$dist (P, π) = \frac{∣ 4 + 6 + 2 - 3 ∣}{3} = \frac{9}{3} = 3$

d) The line through $P$ perpendicular to $π$ is $(4 + t, 3 + 2 t, - 1 - 2 t)$ . Substituting into the plane: $(4 + t) + 2 (3 + 2 t) - 2 (- 1 - 2 t) = 3$ , that is, $12 + 9 t = 3$ and $t = - 1$ . The closest point is

$(3, 1, 1)$

Check: $3 + 2 - 2 = 3$ , and $∣ P M ∣ = ∣(- 1, - 2, 2)∣ = 3$ , which matches the distance in part c).

Example 3 Hard

Two skew lines: common perpendicular and distance

Consider the lines $ℓ_{1} : r = 1 - 1 2 + λ 122$ and $ℓ_{2} : r = 021 + μ 110$ .

a) Show that the lines are skew, and find a vector perpendicular to both, with integer components with no common factor and a positive first component.

b) Find the angle between the two lines, in degrees.

c) Calculate the distance between $ℓ_{1}$ and $ℓ_{2}$ .

d) Find the Cartesian equation of the plane that contains $ℓ_{1}$ and is parallel to $ℓ_{2}$ , in the form $a x + b y + cz + k = 0$ with $a = 2$ , and check the result of part c) using the distance from a point of $ℓ_{2}$ to that plane.

Solution

a) The direction vectors $(1, 2, 2)$ and $(1, 1, 0)$ are not proportional, so the lines are not parallel. To see whether they intersect, different parameters are used:

$⎩ ⎨ ⎧ 1 + λ - 1 + 2 λ 2 + 2 λ = μ = 2 + μ = 1$

The third equation gives $λ = - \frac{1}{2}$ , the first then gives $μ = \frac{1}{2}$ , and the second becomes $- 2 = \frac{5}{2}$ , which is false. The system has no solution: the lines do not intersect and, since they are not parallel, they are skew.

A vector perpendicular to both is

$d_{1} \times d_{2} = 2 \cdot 0 - 2 \cdot 1 2 \cdot 1 - 1 \cdot 0 1 \cdot 1 - 2 \cdot 1 = - 2 2 - 1$

Changing the sign, the required condition is met: $n = (2, - 2, 1)$ . Check: $n \cdot d_{1} = 2 - 4 + 2 = 0$ and $n \cdot d_{2} = 2 - 2 + 0 = 0$ .

b) $d_{1} \cdot d_{2} = 1 + 2 + 0 = 3$ , $∣ d_{1} ∣ = 3$ and $∣ d_{2} ∣ = 2$ :

$cos θ = \frac{3}{3 2} = \frac{1}{2} ⟹ θ = 4 5^{\circ}$

c) With $A_{1} (1, - 1, 2)$ and $A_{2} (0, 2, 1)$ , we have $A_{1} A_{2} = (- 1, 3, - 1)$ . Then

$A_{1} A_{2} \cdot n = - 2 - 6 - 1 = - 9, ∣ n ∣ = 4 + 4 + 1 = 3$

$dist (ℓ_{1}, ℓ_{2}) = \frac{∣ - 9 ∣}{3} = 3$

d) The plane passes through $A_{1} (1, - 1, 2)$ with normal vector $n = (2, - 2, 1)$ : $2 x - 2 y + z = 2 + 2 + 2 = 6$ , that is,

$2 x - 2 y + z - 6 = 0$

The plane contains $ℓ_{1}$ (it passes through $A_{1}$ and its normal is perpendicular to $d_{1}$ ) and is parallel to $ℓ_{2}$ (its normal is perpendicular to $d_{2}$ ). The distance from $A_{2} (0, 2, 1)$ to the plane is

$\frac{∣ 0 - 4 + 1 - 6 ∣}{4 + 4 + 1} = \frac{9}{3} = 3$

the same as in part c).

In the exam: in a part that says "show that the lines are skew", you must prove both things, that they are not parallel and that the system has no solution; one alone is not enough. For the distance, write down which vector you use as the common perpendicular and which vector joins the points: the absolute value in the dot product is what stops a distance from coming out negative.

Common mistakes

Summary

Angle between lines: $cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣ ∣ d _{2} ∣}$ ; between planes: the same with normal vectors. Always between $0^{\circ}$ and $9 0^{\circ}$ .
Angle line–plane: $sin φ = \frac{∣ d \cdot n ∣}{∣ d ∣ ∣ n ∣}$ (sine, not cosine).
Point to line: foot of the perpendicular ( $P Q \cdot d = 0$ with $Q = A + λ d$ ), or $\frac{∣ A P \times d ∣}{∣ d ∣}$ .
Point to plane $a x + b y + cz = d$ : $\frac{∣ a x _{0} + b y _{0} + c z _{0} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$ . The foot is found with the line through $P$ with direction $n$ .
Skew lines: $\frac{∣ A _{1} A _{2} \cdot ( d _{1} \times d _{2} )∣}{∣ d _{1} \times d _{2} ∣}$ , or the distance from $A_{2}$ to the plane that contains $ℓ_{1}$ and is parallel to $ℓ_{2}$ .
Parallel lines or planes: distance from a point of one to the other; line parallel to a plane: distance from a point on the line to the plane.

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