WAEC 2010 · Paper 2 · Q16

Vectors

The position vectors of two points $P$ and $Q$ are $\mathbf p = 5\mathbf i + 3\mathbf j$ and $\mathbf q = 4\mathbf i + 7\mathbf j$ . Find:

(a)
$|6\mathbf p - 5\mathbf q|$ ;
(b)
the scalars $m$ and $n$ such that $m\mathbf p + n\mathbf q = 14\mathbf i + 13\mathbf j$ ;
Separate values with commas, e.g. 3, −2
(c)
the acute angle between $\mathbf p$ and $\mathbf q$ .

Worked solution (try it first)

(a)

$6\mathbf p - 5\mathbf q = (30\mathbf i + 18\mathbf j) - (20\mathbf i + 35\mathbf j)$
$= 10\mathbf i - 17\mathbf j$ .
Magnitude: $\sqrt{10^2 + (-17)^2} = \sqrt{389}$
$\approx 19.723$ .

(b)

$m(5\mathbf i + 3\mathbf j) + n(4\mathbf i + 7\mathbf j) = 14\mathbf i + 13\mathbf j$ .
Compare the $\mathbf i$ and $\mathbf j$ parts: $5m + 4n = 14$ and $3m + 7n = 13$ .
Solve: multiply the first by 3 and the second by 5 and subtract: $23n = 23$ , so $n = 1$ and $m = 2$ .

(c)

$\mathbf p \cdot \mathbf q = 5 \times 4 + 3 \times 7$
$= 41$ , $|\mathbf p| = \sqrt{34}$ , $|\mathbf q| = \sqrt{65}$ .
$\cos\theta = \dfrac{41}{\sqrt{34}\sqrt{65}}$
$= 0.8710$ .
So $\theta = 29.29^\circ$ .

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