WAEC 2023 · Paper 2 · Q15

Vectors

The position vectors of points $P$ , $Q$ and $R$ with respect to the origin are $(-5\mathbf i + \mathbf j)$ , $(2\mathbf i - 4\mathbf j)$ and $(4\mathbf i + 3\mathbf j)$ respectively. If $PQRS$ is a parallelogram, find the:

(a)
position vector of $S$ ;
Show the answer
$-3\mathbf i + 8\mathbf j$
(b)
angle between $\overrightarrow{PS}$ and $\overrightarrow{PQ}$ , correct to one decimal place.

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Worked solution (try it first)

(a)

In $PQRS$ , $\overrightarrow{PS} = \overrightarrow{QR}$
$= (4 - 2)\mathbf i + (3 + 4)\mathbf j$
$= 2\mathbf i + 7\mathbf j$ .
So $\mathbf s = \mathbf p + \overrightarrow{PS}$
$= (-5 + 2)\mathbf i + (1 + 7)\mathbf j$
$= -3\mathbf i + 8\mathbf j$ .

(b)

$\overrightarrow{PQ} = (2 + 5)\mathbf i + (-4 - 1)\mathbf j$
$= 7\mathbf i - 5\mathbf j$ .
$\overrightarrow{PS} \cdot \overrightarrow{PQ} = 14 - 35$
$= -21$ , $|\overrightarrow{PS}| = \sqrt{53}$ and $|\overrightarrow{PQ}| = \sqrt{74}$ .
$\cos\theta = \dfrac{-21}{\sqrt{53}\sqrt{74}}$
$\approx -0.3353$ , so $\theta \approx 109.6^\circ$ .

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