The scalar product · Vectors

Adding vectors gives another vector. The scalar product (or dot product) of two vectors gives a number, and that number tells you the angle between them.

The scalar product and the angle

For $\mathbf a = a_1\mathbf i + a_2\mathbf j$ and $\mathbf b = b_1\mathbf i + b_2\mathbf j$ :

\mathbf a \cdot \mathbf b = a_1b_1 + a_2b_2 = |\mathbf a||\mathbf b|\cos\theta

where $\theta$ is the angle between them when they start from the same point. So

The angle between two vectorscos θ = a·b ÷ (|a||b|)

If $\mathbf a \cdot \mathbf b = 0$ , then $\cos\theta = 0$ : the vectors are perpendicular.
A positive scalar product means an acute angle; a negative one means obtuse.

The scalar productSet the components of a and b

11a·b4.123, 3.606|a| and |b|42.3°angle θ

a₁ = 4a₂ = 1b₁ = 2b₂ = 3

a·b = (4)(2) + (1)(3) = 11. cos θ = 11 ÷ (4.123 × 3.606) = 0.7399, so θ = 42.3°. A positive scalar product means the angle is acute.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q14

The position vectors of points $A$ , $B$ and $C$ with respect to the origin are $(8\mathbf i - 10\mathbf j)$ , $(2\mathbf i + 6\mathbf j)$ and $(-10\mathbf i + 4\mathbf j)$ respectively. If $ABCN$ is a parallelogram, find:

the position vector of $N$ ;

$|\overrightarrow{AN}|$ and $|\overrightarrow{AB}|$ ;

correct to two decimal places, the acute angle between $\overrightarrow{AN}$ and $\overrightarrow{AB}$ .

The fourth vertex
- The $\mathbf i$ part: ${8 + (-10) - 2 = -4}$ .
- The $\mathbf j$ part: ${-10 + 4 - 6 = -12}$ .
- So ${\mathbf n = -4\mathbf i - 12\mathbf j}$ .
Think first. ABCN in order: n = a + c − b.
The two sides from A
- ${\overrightarrow{AN} = \mathbf n - \mathbf a = -12\mathbf i - 2\mathbf j}$ , so ${|\overrightarrow{AN}| = \sqrt{148}}$ .
- ${\overrightarrow{AB} = \mathbf b - \mathbf a = -6\mathbf i + 16\mathbf j}$ , so ${|\overrightarrow{AB}| = \sqrt{292}}$ .
The angle
- ${\overrightarrow{AN} \cdot \overrightarrow{AB} = (-12)(-6) + (-2)(16)}$ .
- ${= 72 - 32 = 40}$ .
- ${\cos\theta = \frac{40}{\sqrt{148}\sqrt{292}} = \frac{40}{207.88} \approx 0.1924}$ .
- So ${\theta \approx 78.91^\circ}$ .
Think first. Find the scalar product, then cos θ.

More: angles between vectors

Perpendicular vectors

To show two vectors are perpendicular, show their scalar product is 0. To find an unknown that makes them perpendicular, set the scalar product equal to 0 and solve.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q8

The points $M$ , $N$ , $Q$ and $R$ are in the $xy$ plane with position vectors $\mathbf m = \mathbf i + \mathbf j$ , $\mathbf n = 2\mathbf i - \mathbf j$ , $\mathbf q = x\mathbf i + \mathbf j$ and $\mathbf r = (x + 1)\mathbf i - 3\mathbf j$ respectively. If $\overrightarrow{NQ}$ is perpendicular to $\overrightarrow{MR}$ , find the values of $x$ .

The two vectors
- ${\overrightarrow{NQ} = \mathbf q - \mathbf n = (x - 2)\mathbf i + 2\mathbf j}$ .
- ${\overrightarrow{MR} = \mathbf r - \mathbf m = x\mathbf i - 4\mathbf j}$ .
Think first. End minus start for NQ and MR.
Scalar product zero
- ${x(x - 2) + 2(-4) = 0}$ .
- So ${x^2 - 2x - 8 = 0}$ .
- Factorise: ${(x - 4)(x + 2) = 0}$ , so ${x = 4}$ or ${x = -2}$ .
Think first. Multiply the i parts, multiply the j parts, add.

More: perpendicular vectors

The scalar product from lengths

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q15 (b)

The vectors $\mathbf a$ and $\mathbf b$ are such that $|\mathbf a| = 3\text{ cm}$ , $|\mathbf b| = 10\text{ cm}$ and $|\mathbf a + \mathbf b| = \sqrt{139}$ . Find: (i) the angle between $\mathbf a$ and $\mathbf b$ ; (ii) the scalar (dot) product of $\mathbf a$ and $\mathbf b$ .

Square the length
- ${|\mathbf a + \mathbf b|^2 = |\mathbf a|^2 + |\mathbf b|^2 + 2\,\mathbf a \cdot \mathbf b}$ .
- ${139 = 9 + 100 + 2\,\mathbf a \cdot \mathbf b}$ , so ${\mathbf a \cdot \mathbf b = 15}$ .
Think first. Expand |a + b|².
The angle
- ${\cos\theta = \frac{15}{3 \times 10} = \frac12}$ , so ${\theta = 60^\circ}$ .

More: unknowns from an angle

Your turn

WAEC 2018 · Paper 2 · Q7

Vectors

(a)
A triangle $PQR$ has vertices $P(2, 2)$ , $Q(3, -1)$ and $R(4, 0)$ . Using the vector method, calculate angle $PQR$ .

Worked solution (try it first)

Angle $PQR$ is at $Q$ , so use vectors starting at $Q$ .
$\overrightarrow{QP} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ and $\overrightarrow{QR} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .
Their scalar product is $(-1)(1) + (3)(1) = 2$ .
$\cos\theta = \dfrac{2}{\sqrt{10}\sqrt2}$
$= \dfrac{1}{\sqrt5}$ , so $\theta \approx 63.43^\circ$ .

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