The Scalar Product

A2 · P3 · 4 min

The scalar (dot) product multiplies two vectors to give a number, and that number encodes the angle between them. It answers "what is the angle?", "are these perpendicular?", and "where is the closest point?" in a few lines each.

Definition and the angle formula

Key result

\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta

where $\theta$ is the angle between the vectors. Hence

\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|},

and $\mathbf{a} \cdot \mathbf{b} = 0$ exactly when the (non-zero) vectors are perpendicular. Also $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$ .

Angle between two vectors

Find the angle between $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$ .

Solution

$\mathbf{a} \cdot \mathbf{b} = 2 - 2 - 4 = -4$ ; $|\mathbf{a}| = 3$ , $|\mathbf{b}| = 3$ .

$\cos\theta = -\tfrac{4}{9}$ , so $\theta = 116.4^\circ$ .

Perpendicular vectors

Find the value of $k$ for which $\begin{pmatrix} k \\ 2 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ k \\ 4 \end{pmatrix}$ are perpendicular.

Solution

$3k + 2k - 4 = 0 \Rightarrow k = \tfrac{4}{5}$ .

Angle between two lines

The angle between lines is the angle between their direction vectors. If the formula gives an obtuse angle, the acute angle $180^\circ - \theta$ is usually what is wanted; state which you are giving.

Angle between lines

Find the acute angle between $\mathbf{r} = \mathbf{a} + s\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\mathbf{r} = \mathbf{c} + t\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ .

Solution

$\cos\theta = \dfrac{2 - 1 + 0}{\sqrt{2}\sqrt{9}} = \dfrac{1}{3\sqrt{2}}$ , so $\theta = 76.4^\circ$ .

Angle in a triangle

For the angle at $B$ in triangle $ABC$ , use $\overrightarrow{BA}$ and $\overrightarrow{BC}$ : both vectors must start at the vertex.

Angle at a vertex

$A(1, 2, 3)$ , $B(4, 2, 7)$ , $C(4, 5, 3)$ . Find angle $ABC$ .

Solution

$\overrightarrow{BA} = \begin{pmatrix} -3 \\ 0 \\ -4 \end{pmatrix}$ , $\overrightarrow{BC} = \begin{pmatrix} 0 \\ 3 \\ -4 \end{pmatrix}$ .

$\cos B = \dfrac{0 + 0 + 16}{5 \times 5} = \dfrac{16}{25}$ , so $B = 50.2^\circ$ .

Foot of the perpendicular from a point to a line

Method

Write the general point on the line, $P$ , in terms of the parameter $t$ .
Form the vector from the given point $Q$ to $P$ : $\overrightarrow{QP} = \mathbf{p} - \mathbf{q}$ .
At the foot of the perpendicular, $\overrightarrow{QP}$ is perpendicular to the direction vector: set $\overrightarrow{QP} \cdot \mathbf{b} = 0$ and solve for $t$ .
Substitute back for the foot; the distance from $Q$ to the line is $|\overrightarrow{QP}|$ at that $t$ .

Shortest distance from a point to a line

Find the foot of the perpendicular from $Q(3, 1, -2)$ to the line $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ , and the distance from $Q$ to the line.

Solution

General point $P = (1 + 2t, -t, 2 + t)$ . $\overrightarrow{QP} = \begin{pmatrix} 2t - 2 \\ -t - 1 \\ t + 4 \end{pmatrix}$ .

Perpendicular to the direction: $2(2t - 2) - (-t - 1) + (t + 4) = 0 \Rightarrow 4t - 4 + t + 1 + t + 4 = 0 \Rightarrow 6t + 1 = 0 \Rightarrow t = -\tfrac{1}{6}$ .

Foot: $\left(\tfrac{2}{3}, \tfrac{1}{6}, \tfrac{11}{6}\right)$ . $\overrightarrow{QP} = \begin{pmatrix} -\tfrac{7}{3} \\ -\tfrac{5}{6} \\ \tfrac{23}{6} \end{pmatrix}$ , distance $= \sqrt{\tfrac{196}{36} + \tfrac{25}{36} + \tfrac{529}{36}} = \sqrt{\tfrac{750}{36}} = \tfrac{5\sqrt{30}}{6} \approx 4.56$ .

Problems with solids

Cuboids, pyramids and prisms are set up with the origin at a corner and axes along edges. Write each vertex as a position vector, then everything above applies.

Angle in a cuboid

A cuboid has $O$ at one corner, $A(6, 0, 0)$ , $C(0, 4, 0)$ , $D(0, 0, 3)$ along the edges, and $G$ the vertex opposite $O$ . Find the angle between the diagonal $OG$ and the edge $OA$ .

Solution

$\mathbf{g} = \begin{pmatrix} 6 \\ 4 \\ 3 \end{pmatrix}$ , $\mathbf{a} = \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix}$ . $\cos\theta = \dfrac{36}{\sqrt{61} \times 6} = \dfrac{6}{\sqrt{61}}$ , so $\theta = 39.8^\circ$ .

Watch out

The scalar product of two vectors is a number. If your answer to $\mathbf{a} \cdot \mathbf{b}$ is a vector, you have multiplied component-wise without adding.

Exam tip

When the question gives an angle and asks for a value of an unknown, you will get a quadratic from $\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ after squaring. Squaring can introduce a false root with the wrong sign of the dot product, so check both candidates.

Practice

Question

Find the angle between $3\mathbf{i} + 4\mathbf{k}$ and $\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$ .
Show that the lines $\mathbf{r} = \mathbf{a} + s(\mathbf{i} + 2\mathbf{j} - \mathbf{k})$ and $\mathbf{r} = \mathbf{c} + t(3\mathbf{i} - \mathbf{j} + \mathbf{k})$ are perpendicular.
Find the foot of the perpendicular from the origin to the line $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} + t\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$ .
$P(2, 1, 0)$ , $Q(4, 3, -1)$ , $R(1, 3, 2)$ . Find angle $PQR$ .

Answers

$\cos\theta = \dfrac{3 + 8}{5 \times 3} = \tfrac{11}{15}$ , $\theta = 42.8^\circ$ .
$3 - 2 - 1 = 0$ .
$P = (1 + t, 2, 2 - t)$ ; $\mathbf{p} \cdot \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = 1 + t - 2 + t = 0 \Rightarrow t = \tfrac{1}{2}$ ; foot $\left(\tfrac{3}{2}, 2, \tfrac{3}{2}\right)$ .
$\overrightarrow{QP} = \begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}$ , $\overrightarrow{QR} = \begin{pmatrix} -3 \\ 0 \\ 3 \end{pmatrix}$ ; $\cos Q = \dfrac{6 + 3}{3 \times 3\sqrt{2}} = \tfrac{1}{\sqrt{2}}$ , $Q = 45^\circ$ .

Definition and the angle formula

Angle between two lines

Angle in a triangle

Foot of the perpendicular from a point to a line

Problems with solids

Practice

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