WAEC 2017 · Paper 2 · Q15

(a)
Given that $\mathbf m = 6\mathbf i + 8\mathbf j$ and $\mathbf n = -8\mathbf i + \frac73\mathbf j$ , find the: (i) magnitudes and directions (bearings) of $\mathbf m$ and $\mathbf n$ ; (ii) angle between $\mathbf m$ and $\mathbf n$ .
(b)
The position vectors of points $P$ , $Q$ , $R$ , $S$ are $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ , $\begin{pmatrix} 10 \\ 4 \end{pmatrix}$ , $\begin{pmatrix} 3 \\ 12 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ respectively. Show that $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{RS}$ .
Show the answer
$\overrightarrow{PQ} \cdot \overrightarrow{RS} = 12 - 12 = 0$

Worked solution (try it first)

$|\mathbf m| = 10$ .
$\mathbf m$ points 6 east and 8 north, a bearing of $\tan^{-1}\frac68 \approx 037^\circ$ .
$|\mathbf n| = \sqrt{64 + \frac{49}{9}}$
$= \frac{25}{3}$ .
$\mathbf n$ points 8 west and $\frac73$ north.
So its bearing is $360^\circ - \tan^{-1}\frac{24}{7} = 360^\circ - 73.7^\circ$
$\approx 286^\circ$ .

$\mathbf m \cdot \mathbf n = -48 + \frac{56}{3}$
$= -\frac{88}{3}$ .
$\cos\theta = \dfrac{-88/3}{10 \times 25/3}$
$= -\dfrac{88}{250}$
$= -0.352$ , so $\theta \approx 110.6^\circ$ .

$\overrightarrow{PQ} = \begin{pmatrix} 12 \\ 1 \end{pmatrix}$ and $\overrightarrow{RS} = \begin{pmatrix} 1 \\ -12 \end{pmatrix}$ .
$\overrightarrow{PQ} \cdot \overrightarrow{RS} = 12 - 12$
$= 0$ , so they are perpendicular.