WAEC 2014 · Paper 2 · Q14

Vectors

The position vectors of two points $P$ and $R$ are $\mathbf p = 4\mathbf i - 2\mathbf j$ and $\mathbf r = 2\mathbf i + 3\mathbf j$ respectively. Find:

(a)
$|3\mathbf p - 2\mathbf r|$ ;
(b)
the values of the scalars $m$ and $n$ such that $8\mathbf i + 8\mathbf j = m\mathbf p + n\mathbf r$ ;
Separate values with commas, e.g. 3, −2
(c)
the cosine of the acute angle between $\mathbf p$ and $\mathbf r$ , leaving the answer in surd form.

Worked solution (try it first)

(a)

$3\mathbf p - 2\mathbf r = (12 - 4)\mathbf i + (-6 - 6)\mathbf j$
$= 8\mathbf i - 12\mathbf j$ .
$|8\mathbf i - 12\mathbf j| = \sqrt{64 + 144}$
$= \sqrt{208}$
$= 4\sqrt{13}$ .

(b)

Match the parts: $4m + 2n = 8$ and $-2m + 3n = 8$ .
Double the second and add: $8n = 24$ , so $n = 3$ .
Then $4m = 8 - 6$ , so $m = \frac12$ .

(c)

$\mathbf p \cdot \mathbf r = 8 - 6 = 2$ , $|\mathbf p| = \sqrt{20}$ and $|\mathbf r| = \sqrt{13}$ .
$\cos\theta = \dfrac{2}{\sqrt{260}}$
$= \dfrac{2}{2\sqrt{65}}$
$= \dfrac{1}{\sqrt{65}}$
$= \dfrac{\sqrt{65}}{65}$ .

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