WAEC 2018 · Paper 2 · Q14

Vectors

The position vectors of points $A$ , $B$ and $C$ with respect to the origin are $(8\mathbf i - 10\mathbf j)$ , $(2\mathbf i + 6\mathbf j)$ and $(-10\mathbf i + 4\mathbf j)$ respectively. If $ABCN$ is a parallelogram, find:

(a)
the position vector of $N$ ;
Show the answer
$-4\mathbf i - 12\mathbf j$
(b)
$|\overrightarrow{AN}|$ and $|\overrightarrow{AB}|$ ;
Separate values with commas, e.g. 3, −2
(c)
correct to two decimal places, the acute angle between $\overrightarrow{AN}$ and $\overrightarrow{AB}$ .

Worked solution (try it first)

(a)

In $ABCN$ , $\overrightarrow{AB} = \overrightarrow{NC}$ , so $\mathbf n = \mathbf a + \mathbf c - \mathbf b$ .
$\mathbf n = (8 - 10 - 2)\mathbf i + (-10 + 4 - 6)\mathbf j$
$= -4\mathbf i - 12\mathbf j$ .

(b)

$\overrightarrow{AN} = \mathbf n - \mathbf a$
$= -12\mathbf i - 2\mathbf j$ , so $|\overrightarrow{AN}| = \sqrt{148}$
$\approx 12.166$ .
$\overrightarrow{AB} = \mathbf b - \mathbf a$
$= -6\mathbf i + 16\mathbf j$ , so $|\overrightarrow{AB}| = \sqrt{292}$
$\approx 17.088$ .

(c)

$\overrightarrow{AN} \cdot \overrightarrow{AB} = 72 - 32$
$= 40$ .
$\cos\theta = \dfrac{40}{\sqrt{148}\sqrt{292}}$
$\approx 0.1924$ , so $\theta \approx 78.91^\circ$ .

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