WAEC 2014 · Paper 2 · Q5

Vectors

The position vectors of points $P$ , $Q$ and $R$ are $11\mathbf i + \mathbf j$ , $5\mathbf i + \frac{13}{3}\mathbf j$ and $2\mathbf i + 6\mathbf j$ respectively.

(a)
Show that $P$ , $Q$ and $R$ lie on a straight line.
Show the answer
$\overrightarrow{PQ} = 2\overrightarrow{QR}$ , so $P$ , $Q$ , $R$ are collinear
(b)
Find the ratio $|\overrightarrow{PQ}| : |\overrightarrow{QR}|$ .

Worked solution (try it first)

(a)

$\overrightarrow{PQ} = \mathbf q - \mathbf p$ , which is $(5 - 11)\mathbf i + \left(\frac{13}{3} - 1\right)\mathbf j = -6\mathbf i + \frac{10}{3}\mathbf j$ .
$\overrightarrow{QR} = \mathbf r - \mathbf q$ , which is $(2 - 5)\mathbf i + \left(6 - \frac{13}{3}\right)\mathbf j = -3\mathbf i + \frac53\mathbf j$ .
So $\overrightarrow{PQ} = 2\overrightarrow{QR}$ : the two vectors are parallel.
They share the point $Q$ , so $P$ , $Q$ and $R$ lie on a straight line.

(b)

$|\overrightarrow{PQ}| = \sqrt{36 + \frac{100}{9}}$
$= \frac{2\sqrt{106}}{3}$ .
$|\overrightarrow{QR}| = \sqrt{9 + \frac{25}{9}}$
$= \frac{\sqrt{106}}{3}$ .
So $|\overrightarrow{PQ}| : |\overrightarrow{QR}| = 2 : 1$ .

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