Further Maths/Lessons/Vectors/The Scalar Product & Angles
Core Pure · Vectors · Year 1

The Scalar Product & Angles

The scalar product turns two vectors into a single number — and that number tells you the angle between them.

13 min read 3 worked examples

The scalar (dot) product

Definition 2.1 — scalar product

For $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$: $$\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3=|\mathbf{a}||\mathbf{b}|\cos\theta$$ where $\theta$ is the angle between the vectors when placed tail to tail.

Theorem 2.2 — angle between two vectors

$$\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$$ Two non-zero vectors are perpendicular exactly when $\mathbf{a}\cdot\mathbf{b}=0$.

Two vectors a and b drawn from the same starting point with the angle theta between them marked with an arc, illustrating the geometric meaning of the scalar product
Fig. 1 — the angle θ between vectors a and b, measured where the two vectors are placed tail to tail.
Example 1 · Finding the angle between two vectors
Find the angle between $\mathbf{a}=(2,1,-2)$ and $\mathbf{b}=(1,3,4)$, to 1 decimal place.

$\mathbf{a}\cdot\mathbf{b}=2(1)+1(3)+(-2)(4)=2+3-8=-3$.

$|\mathbf{a}|=\sqrt{4+1+4}=3$, $|\mathbf{b}|=\sqrt{1+9+16}=\sqrt{26}$.

$$\cos\theta=\frac{-3}{3\sqrt{26}}=-0.19612\ldots \Rightarrow \theta=101.3°$$
Example 2 · Finding an unknown for perpendicular vectors
Find $k$ such that $\mathbf{a}=(3,k,-1)$ and $\mathbf{b}=(2,-1,5)$ are perpendicular.

$\mathbf{a}\cdot\mathbf{b}=3(2)+k(-1)+(-1)(5)=6-k-5=1-k$.

Perpendicular $\Rightarrow 1-k=0 \Rightarrow k=1$.

Example 3 · Angle between two lines
Find the acute angle between the lines with direction vectors $(1,2,2)$ and $(2,-2,1)$.

Dot product: $1(2)+2(-2)+2(1)=2-4+2=0$.

The direction vectors are perpendicular, so the lines meet (or would meet if extended) at $90°$.

Exam technique

If a calculated angle from $\cos\theta$ comes out obtuse but the question asks for the angle between two lines (rather than two directed vectors), take the acute angle instead by using $|\cos\theta|$ — lines don't have a "direction", so the angle between them is always taken to be $\leq90°$.

Exam-style questions

Question 1 5 marks

Find the angle between $\mathbf{a}=(1,-2,2)$ and $\mathbf{b}=(2,2,1)$, giving your answer to 1 decimal place.

$\mathbf{a}\cdot\mathbf{b}=1(2)+(-2)(2)+2(1)=2-4+2=0$ (M1, A1)

Since the scalar product is $0$, the vectors are perpendicular: $\theta=90°$. (A1, B1, B1)

Question 2 6 marks

Given $\mathbf{a}=(4,-1,3)$ and $\mathbf{b}=(1,2,-2)$, find $\mathbf{a}\cdot\mathbf{b}$, $|\mathbf{a}|$, $|\mathbf{b}|$, and the angle between $\mathbf{a}$ and $\mathbf{b}$ to 1 decimal place.

$\mathbf{a}\cdot\mathbf{b}=4(1)+(-1)(2)+3(-2)=4-2-6=-4$ (M1, A1)

$|\mathbf{a}|=\sqrt{16+1+9}=\sqrt{26}$, $|\mathbf{b}|=\sqrt{1+4+4}=3$ (A1)

$\cos\theta=\dfrac{-4}{3\sqrt{26}}=-0.2614\ldots \Rightarrow \theta=105.2°$ (M1, A1, A1)

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