IB Analysis and Approaches HL · Plane geometry: vectors and lines · Section 2 of 3

Scalar (dot) product and angle between vectors

3 worked examples · By Álvaro López

Key idea

The scalar (dot) product turns two vectors into a number that measures how much they "point in the same direction". With it you calculate the angle between two vectors, check whether they are perpendicular, and find the angle between two lines. It appears in Paper 1 with simple vectors and in Paper 2 within geometry and motion problems.

Two ways to calculate it

Given two vectors $u$ and $v$ with the same origin, let $θ$ be the angle between them, with $0^{\circ} \leq θ \leq 18 0^{\circ}$ . For $θ$ to be well defined, the two vectors are drawn starting from the same point.

Two vectors u and v with the same origin O and angle theta between them; the scalar projection of v onto u measures the magnitude of v times the cosine of theta

Why the two formulas give the same result

The vectors $u$ , $v$ and $u - v$ form a triangle whose sides measure $∣ u ∣$ , $∣ v ∣$ and $∣ u - v ∣$ , and the angle between the first two is $θ$ . The cosine rule states:

$∣ u - v ∣^{2} = ∣ u ∣^{2} + ∣ v ∣^{2} - 2 ∣ u ∣ ∣ v ∣ cos θ$

On the other hand, in components, $∣ u - v ∣^{2} = (u_{1} - v_{1})^{2} + (u_{2} - v_{2})^{2}$ . Expanding and simplifying the squares $u_{1}^{2} + u_{2}^{2} = ∣ u ∣^{2}$ and $v_{1}^{2} + v_{2}^{2} = ∣ v ∣^{2}$ , we get

$- 2 u_{1} v_{1} - 2 u_{2} v_{2} = - 2 ∣ u ∣ ∣ v ∣ cos θ ⟹ u_{1} v_{1} + u_{2} v_{2} = ∣ u ∣ ∣ v ∣ cos θ$

The cosine form is useful when you know magnitudes and the angle; the component form is useful when you know the coordinates. They are almost always used together: one is equated with the other to find the angle or a parameter.

Properties

The penultimate property comes from the formula with $θ = 0^{\circ}$ : a vector makes an angle of $0^{\circ}$ with itself, and $cos 0^{\circ} = 1$ . The others can be checked in a moment with components. What there is not is a "division" or an associative property: $(u \cdot v) \cdot w$ is meaningless, because $u \cdot v$ is already a number.

The sign of the result tells you the type of angle, because the only thing in $∣ u ∣ ∣ v ∣ cos θ$ that can change sign is the cosine (the magnitudes are positive if the vectors are not zero):

Angle $θ$	$cos θ$	$u \cdot v$
$0^{\circ} \leq θ < 9 0^{\circ}$ (acute)	positive	$> 0$
$θ = 9 0^{\circ}$ (right)	zero	$= 0$
$9 0^{\circ} < θ \leq 18 0^{\circ}$ (obtuse)	negative	$< 0$

Angle between two vectors

Solving for $cos θ$ in the geometric formula and substituting the component form:

As the cosine is one-to-one between $0^{\circ}$ and $18 0^{\circ}$ , there is no ambiguous case here: the angle is unique, and its sign tells you whether it is acute or obtuse. If the angle the problem asks for is that of a triangle $A B C$ at the vertex $B$ , use the vectors $B A$ and $B C$ , both starting from $B$ .

Perpendicular vectors

If $θ = 9 0^{\circ}$ then $cos θ = 0$ , and conversely. For non-zero vectors:

A useful consequence: in the plane, the vector $(- b a)$ is perpendicular to $(a b)$ , because $a (- b) + b a = 0$ . Simply swap the components and change the sign of one of them. If you need a vector with a perpendicularity condition that depends on a parameter, set up $u \cdot v = 0$ and solve it.

Scalar projection

Look at the figure again. If you drop a perpendicular from the end of $v$ onto the line of $u$ , the segment it leaves on that line has length $∣ v ∣ cos θ$ : it is the "shadow" of $v$ on the direction of $u$ .

This quantity is the signed distance that $v$ advances in the direction of $u$ . The distance from a point to a line is based on the same idea.

Angle between two lines

Two lines that intersect form four angles, equal in pairs: an acute angle $α$ (or a right angle) and its supplement $18 0^{\circ} - α$ . By convention, the angle between two lines is the acute one (or $9 0^{\circ}$ ). It is calculated with their direction vectors $u$ and $v$ , which may form an acute angle or its obtuse supplement depending on the sense in which you wrote them. To avoid depending on that sense, the absolute value is taken:

If a line is given in Cartesian form $a x + b y + c = 0$ , a direction vector is $(b - a)$ . In Paper 2, the angle between two lines with gradients $m_{1}$ and $m_{2}$ can also be obtained with the GDC, but the vector method is the one you are asked to justify.

Worked examples

Example 1 Basic

Scalar product using magnitudes and using components

a) The vectors $p$ and $q$ satisfy $∣ p ∣ = 4$ , $∣ q ∣ = 5$ and form an angle of $6 0^{\circ}$ . Calculate $p \cdot q$ .

b) Calculate $u \cdot v$ if $u = (2 - 3)$ and $v = (45)$ , and state whether the angle between them is acute, right or obtuse.

c) Find $k$ so that $(k 3)$ and $(6 - 2)$ are perpendicular.

Solution

a) With the geometric form: $p \cdot q = 4 \cdot 5 \cdot cos 6 0^{\circ} = 20 \cdot \frac{1}{2} = 10$ .

b) In components: $u \cdot v = 2 \cdot 4 + (- 3) \cdot 5 = 8 - 15 = - 7$ . It is negative, so the angle is obtuse.

c) Perpendicular is equivalent to a zero scalar product: $6 k + 3 \cdot (- 2) = 0$ , that is, $6 k - 6 = 0$ , so $k = 1$ .

Example 2 Medium

Angles and projection in a triangle

A triangle has vertices $A (1, 1)$ , $B (0, 4)$ and $C (3, 5)$ .

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

b) Find the angle of the triangle at $A$ (in radians, exact).

c) Calculate $B A \cdot B C$ .

d) Calculate the scalar projection of $A C$ onto $A B$ .

Solution

First the vectors, subtracting end minus start:

$A B = (0 - 1 4 - 1) = (- 1 3), A C = (3 - 1 5 - 1) = (24)$

a) $A B \cdot A C = (- 1) \cdot 2 + 3 \cdot 4 = 10$ .

b) The magnitudes are $∣ A B ∣ = 1 + 9 = 10$ and $∣ A C ∣ = 4 + 16 = 25$ . Then

$cos A = \frac{10}{10 \cdot 2 5} = \frac{10}{2 50} = \frac{10}{10 2} = \frac{2}{2}$

and the angle is $A = \frac{π}{4}$ (that is, $4 5^{\circ}$ ).

c) At $B$ we use vectors that start from $B$ : $B A = (1 - 3)$ and $B C = (31)$ . Their product is $1 \cdot 3 + (- 3) \cdot 1 = 0$ : the angle at $B$ is right.

d) $\frac{A B \cdot A C}{∣ A B ∣} = \frac{10}{10} = 10$ . It coincides with $∣ A B ∣$ , as expected: since the angle at $B$ is right, the foot of the perpendicular from $C$ to the line $A B$ is exactly $B$ .

Example 3 Hard

A parameter and the angle between two paths

a) The vectors $u = (k 1)$ and $v = (13)$ form an angle of $4 5^{\circ}$ . Find all the possible values of $k$ .

b) Two straight paths in a park have direction vectors $v = (13)$ and $w = (- 2 - 1)$ . Find the angle between the vectors $v$ and $w$ (in radians, exact).

c) Find the acute angle between the two paths (in radians, exact).

Solution

a) Equate $cos 4 5^{\circ} = \frac{2}{2}$ with the angle formula. As $u \cdot v = k + 3$ , $∣ u ∣ = k^{2} + 1$ and $∣ v ∣ = 10$ :

$\frac{k + 3}{k ^{2} + 1 10} = \frac{2}{2}$

To remove the root, square both sides (you must check the result afterwards, because the sign is lost):

$\frac{( k + 3 ) ^{2}}{10 ( k ^{2} + 1 )} = \frac{1}{2} ⟹ 2 (k + 3)^{2} = 10 (k^{2} + 1) ⟹ k^{2} + 6 k + 9 = 5 k^{2} + 5$

$4 k^{2} - 6 k - 4 = 0 ⟹ 2 k^{2} - 3 k - 2 = 0 ⟹ k = \frac{3 \pm 5}{4} ⟹ k = 2 or k = - \frac{1}{2}$

Check: the original cosine is positive, so $k + 3 > 0$ is needed. It holds for $k = 2$ ( $5 > 0$ ) and for $k = - \frac{1}{2}$ ( $\frac{5}{2} > 0$ ): both values are valid.

b) $v \cdot w = 1 \cdot (- 2) + 3 \cdot (- 1) = - 5$ , $∣ v ∣ = 10$ and $∣ w ∣ = 5$ :

$cos θ = \frac{- 5}{10 5} = \frac{- 5}{5 2} = - \frac{2}{2} ⟹ θ = \frac{3 π}{4}$

c) The result of b) is an obtuse angle ( $13 5^{\circ}$ ), but the angle between lines is the acute one, that is, the supplement: $π - \frac{3 π}{4} = \frac{π}{4}$ . It also comes out using the absolute value: $cos α = \frac{∣ - 5 ∣}{5 2} = \frac{2}{2}$ , so $α = \frac{π}{4}$ ( $4 5^{\circ}$ ).

In the exam: when you square, check the sign of the scalar product to avoid false solutions. And when talking about lines, always write the acute angle, even if the scalar product of your vectors comes out negative.

Common mistakes

Summary

Scalar product: $u \cdot v = ∣ u ∣ ∣ v ∣ cos θ = u_{1} v_{1} + u_{2} v_{2}$ . It is a number, not a vector. The two forms agree by the cosine rule.
Properties: commutative, distributive, $(k u) \cdot v = k (u \cdot v)$ , $u \cdot u = ∣ u ∣^{2}$ . Positive if the angle is acute, negative if it is obtuse.
Angle between vectors: $cos θ = \frac{u \cdot v}{∣ u ∣ ∣ v ∣}$ , with both vectors starting from the same point.
Perpendicular: $u \cdot v = 0$ . The vector $(- b a)$ is perpendicular to $(a b)$ .
Scalar projection of $v$ onto $u$ : $∣ v ∣ cos θ = \frac{u \cdot v}{∣ u ∣}$ .
Angle between lines (acute): $cos α = \frac{∣ u \cdot v ∣}{∣ u ∣ ∣ v ∣}$ with their direction vectors.

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