WAEC 2008 · Paper 2 · Q16

Vectors

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

(a)
the position vector of $M$ ;
Separate values with commas, e.g. 3, −2
(b)
$|\overrightarrow{PM}|$ and $|\overrightarrow{PQ}|$ ;
Separate values with commas, e.g. 3, −2
(c)
the acute angle between $\overrightarrow{PM}$ and $\overrightarrow{PQ}$ , correct to one decimal place;
(d)
the area of $PQRM$ .

Worked solution (try it first)

(a)

In parallelogram $PQRM$ , $\overrightarrow{MP} = \overrightarrow{RQ}$ , so $\mathbf p - \mathbf m = \mathbf q - \mathbf r$ .
Rearrange: $\mathbf m = \mathbf p + \mathbf r - \mathbf q$ .
$\mathbf m = (4 - 5 - 1)\mathbf i + (-5 + 2 - 3)\mathbf j$
$= -2\mathbf i - 6\mathbf j$ .

(b)

$\overrightarrow{PM} = \mathbf m - \mathbf p$
$= -6\mathbf i - \mathbf j$ , so $|\overrightarrow{PM}| = \sqrt{36 + 1}$
$= \sqrt{37}$ .
$\overrightarrow{PQ} = \mathbf q - \mathbf p$
$= -3\mathbf i + 8\mathbf j$ , so $|\overrightarrow{PQ}| = \sqrt{9 + 64}$
$= \sqrt{73}$ .

(c)

Dot product: $\overrightarrow{PM} \cdot \overrightarrow{PQ} = (-6)(-3) + (-1)(8)$
$= 10$ .
$\cos\theta = \dfrac{10}{\sqrt{37}\sqrt{73}}$
$\approx 0.1924$ , so $\theta \approx 78.9^\circ$ .

(d)

Area of a parallelogram $= |\overrightarrow{PM}||\overrightarrow{PQ}|\sin\theta$ .
$\sqrt{37}\sqrt{73}\sin\theta = \sqrt{2701} \times 0.9813$
$\approx 51$ square units.
(Exactly: $|(-6)(8) - (-1)(-3)| = 51$ .)

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