WAEC 2018 · Paper 2 · Q7

(a)
Two points $X$ and $Y$ have position vectors $\mathbf x = 2\mathbf i - 3\mathbf j$ and $\mathbf y = -\mathbf i + 2\mathbf j$ . Find the position vector of the point $M$ on $\overline{XY}$ such that $|\overrightarrow{XM}| : |\overrightarrow{MY}| = 3 : 2$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$XM : MY = 3 : 2$ , so $M$ is $\frac35$ of the way from $X$ to $Y$ .
$\overrightarrow{OM} = \dfrac{2\mathbf x + 3\mathbf y}{5}$ .
$= \dfrac{2(2\mathbf i - 3\mathbf j) + 3(-\mathbf i + 2\mathbf j)}{5}$
$= \dfrac{\mathbf i + 0\mathbf j}{5}$ .
So $\overrightarrow{OM} = \frac15\mathbf i$ .