Learn9709Paper 3 · Pure Mathematics 3VectorsThe scalar product and angles

Vectors · Subtopic 2 of 5

The Scalar Product and Angles

Multiply two vectors and get a number that measures how much they point the same way. Use it to find any angle in three dimensions, and to test for right angles in a single line of working.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14AQA 7367 Paper 1≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Calculate $\mathbf{a}\cdot\mathbf{b}$ from components.
Find the angle between two vectors using $\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ .
Use $\mathbf{a}\cdot\mathbf{b}=0$ to show vectors are perpendicular or to find an unknown.
Find angles inside triangles, choosing vectors that point away from the vertex.

Pause and think

$\mathbf{a}=\begin{pmatrix}2\\1\\-2\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}1\\2\\2\end{pmatrix}$ . Multiply matching components and add. What do you get, and what might it mean?

Section 2 of 14

Core idea

Start hereAngles without protractors

How do you measure the angle between two flight paths in 3D? No protractor fits. The scalar product gives it from the components alone, and tells you instantly when two directions are perpendicular.

The big idea

$\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta$ . So $\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}$ ; $\mathbf a\cdot\mathbf b=0\iff$ perpendicular.

1. Why the two formulas agree

a · b = |a||b| cos θ as the angle opens

θ (degrees) = 15

\mathbf a\cdot\mathbf b

is the length of

\mathbf a

times the "shadow" of

\mathbf b

on it. At

90^\circ

the shadow is zero. Past

90^\circ

the shadow points backwards, so the product is negative.

Using the cosine rule

$|\mathbf b-\mathbf a|^2=|\mathbf a|^2+|\mathbf b|^2-2|\mathbf a||\mathbf b|\cos\theta$
Cosine rule in the triangle with sides $\mathbf a$ , $\mathbf b$ , $\mathbf b-\mathbf a$ .
$\sum(b_i-a_i)^2=\sum a_i^2+\sum b_i^2-2\sum a_ib_i$
Expand the left in components.
$\Rightarrow\sum a_ib_i=|\mathbf a||\mathbf b|\cos\theta$
Compare.

Worked exampleAngle between $(1,2,2)$ and $(3,0,4)$

Find it.

Goal: $\theta$ .
Strategy: Dot product over magnitudes.

$\mathbf a\cdot\mathbf b=3+0+8=11$
Component sum.
$|\mathbf a|=3,\ |\mathbf b|=5$
Magnitudes.
$\cos\theta=\tfrac{11}{15}\Rightarrow\theta=42.8^\circ$
Inverse cosine.

Answer

42.8^\circ

✓ Check

\cos\theta>0

, acute ✓.

Worked exampleAngle $ABC$ in a triangle

$A(1,0,2)$ , $B(2,1,0)$ , $C(4,0,1)$ . Find angle $ABC$ .

Goal: Angle at $B$ .
Strategy: Both vectors must start at $B$ .

$\overrightarrow{BA}=(-1,-1,2),\ \overrightarrow{BC}=(2,-1,1)$
From $B$ .
$\cos B=\frac{-2+1+2}{\sqrt6\sqrt6}=\tfrac16$
Formula.

Answer

80.4^\circ

✓ CheckUsing

\overrightarrow{AB}

instead would give the supplement

99.6^\circ

: wrong ✓.

Common misconceptionDirection of vectors For the angle at

B

, use

\overrightarrow{BA}

and

\overrightarrow{BC}

, both leaving

B

Pause and think

Are $(1,1,1)$ and $(1,-2,1)$ perpendicular?

Summary

Two formulas for $\mathbf a\cdot\mathbf b$ .
Zero ⇔ perpendicular.

Scalar product: Dot product; a number.

HistoryA dot and a cross. Josiah Willard Gibbs, a professor at Yale, printed his lecture notes Elements of Vector Analysis privately in 1881 and 1884. He introduced the dot

\mathbf{a}\cdot\mathbf{b}

for the scalar product and the cross

\mathbf{a}\times\mathbf{b}

for the vector product, and the notation has stayed ever since.

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