WAEC 2009 · Paper 2 · Q16✱

(a)
Find the unit vector along the resultant of the vectors $(3\mathbf i - 2\mathbf j)$ and $(\mathbf i + 5\mathbf j)$ .
Separate values with commas, e.g. 3, −2
(b)
The position vectors of points $P$ , $Q$ , $R$ and $S$ 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. (i) Show that $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{RS}$ . (ii) Calculate, correct to one decimal place, angle $PSQ$ .

Worked solution (try it first)

Resultant: $(3\mathbf i - 2\mathbf j) + (\mathbf i + 5\mathbf j) = 4\mathbf i + 3\mathbf j$ .
Its magnitude is $\sqrt{4^2 + 3^2} = 5$ , so the unit vector is $\frac15(4\mathbf i + 3\mathbf j) = \frac45\mathbf i + \frac35\mathbf j$ .

$\overrightarrow{PQ} = \begin{pmatrix} 10 \\ 4 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}$
$= \begin{pmatrix} 12 \\ 1 \end{pmatrix}$ and $\overrightarrow{RS} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ 12 \end{pmatrix}$
$= \begin{pmatrix} 1 \\ -12 \end{pmatrix}$ .
$\overrightarrow{PQ}\cdot\overrightarrow{RS} = 12 \times 1 + 1 \times (-12)$
$= 0$ , so $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{RS}$ .

$\overrightarrow{SP} = \begin{pmatrix} -2 - 4 \\ 3 - 0 \end{pmatrix}$
$= \begin{pmatrix} -6 \\ 3 \end{pmatrix}$ and $\overrightarrow{SQ} = \begin{pmatrix} 10 - 4 \\ 4 - 0 \end{pmatrix}$
$= \begin{pmatrix} 6 \\ 4 \end{pmatrix}$ .
$\overrightarrow{SP}\cdot\overrightarrow{SQ} = -36 + 12$
$= -24$ .
$|\overrightarrow{SP}| = \sqrt{45}$
$= 3\sqrt5$ and $|\overrightarrow{SQ}| = \sqrt{52}$
$= 2\sqrt{13}$ .
$\cos\angle PSQ = \dfrac{-24}{3\sqrt5 \times 2\sqrt{13}}$
$= \dfrac{-4}{\sqrt{65}}$
$= -0.4961$ .
The cosine is negative, so the angle is obtuse: $\angle PSQ = 119.7^\circ$ (1 d.p.).