WAEC 2017 · Paper 2 · Q15

(a)
The position vectors of the points $M$ , $N$ , $P$ relative to a fixed point $O$ are $\mathbf m = 2\mathbf i + 3\mathbf j$ , $\mathbf n = 6\mathbf i + 23\mathbf j$ , $\mathbf p = 2\mathbf i - 7\mathbf j$ . Find $|3\mathbf m + 2\mathbf n - 5\mathbf p|$ (3 d.p.).
(b)
Forces $F_1 = 18\text{ N}$ , $F_2 = 15\text{ N}$ and $F_3 = 20\text{ N}$ act in the directions of $-4\mathbf i$ , $3\mathbf i + 4\mathbf j$ and $8\mathbf j$ respectively. Find, correct to three decimal places, the magnitude and direction of the resultant force.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$3\mathbf m + 2\mathbf n - 5\mathbf p = (6 + 12 - 10)\mathbf i + (9 + 46 + 35)\mathbf j$
$= 8\mathbf i + 90\mathbf j$ .
Its magnitude is $\sqrt{64 + 8100} = \sqrt{8164}$
$\approx 90.355$ .

Each force is its size times the unit vector of its direction.
$F_1 = -18\mathbf i$ and $F_3 = 20\mathbf j$ .
$|3\mathbf i + 4\mathbf j| = 5$ , so $F_2 = 15 \times \frac15(3\mathbf i + 4\mathbf j)$
$= 9\mathbf i + 12\mathbf j$ .
Resultant: $(-18 + 9)\mathbf i + (12 + 20)\mathbf j = -9\mathbf i + 32\mathbf j$ .
$|\mathbf R| = \sqrt{81 + 1024}$
$= \sqrt{1105}$
$\approx 33.242\text{ N}$ .
It points north-west: $\tan^{-1}\frac{9}{32} = 15.71^\circ$ west of north, a bearing of $344.29^\circ$ .