WAEC 2009 · Paper 2 · Q16

(a)
The position vectors of points $L$ and $M$ are $(5\mathbf i + 6\mathbf j)$ and $(13\mathbf i + 4\mathbf j)$ respectively. If point $K$ lies on $LM$ such that $LK : KM = 2 : 3$ , find the position vector of $K$ .
Separate values with commas, e.g. 3, −2
(b)
Three poles are situated at points $A$ , $B$ and $C$ on the same horizontal plane such that $\overrightarrow{AB} = (8\text{ km}, 060^\circ)$ and $\overrightarrow{BC} = (12\text{ km}, 130^\circ)$ . Calculate: (i) $|AC|$ , correct to three decimal places; (ii) the bearing of $C$ from $A$ , correct to the nearest degree.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

By the ratio theorem, $\mathbf k = \dfrac{3\mathbf l + 2\mathbf m}{2 + 3}$ (each end is weighted by the part of the ratio next to the other end).
$3(5\mathbf i + 6\mathbf j) + 2(13\mathbf i + 4\mathbf j) = 41\mathbf i + 26\mathbf j$ .
Divide by 5: $\mathbf k = 8.2\mathbf i + 5.2\mathbf j$ .

At $B$ , the back bearing to $A$ is $060^\circ + 180^\circ = 240^\circ$ .
The angle between $BA$ and $BC$ is $240^\circ - 130^\circ = 110^\circ$ .
Cosine rule: $|AC|^2 = 8^2 + 12^2 - 2(8)(12)\cos 110^\circ$
$= 208 + 65.666$
$= 273.666$ .
So $|AC| = 16.543$ km (3 d.p.).

Sine rule for $\theta = \angle BAC$ : $\dfrac{\sin\theta}{12} = \dfrac{\sin 110^\circ}{16.543}$ , so $\sin\theta = 0.6816$ and $\theta = 42.97^\circ$ .
The bearing of $C$ from $A$ is $060^\circ + 42.97^\circ = 102.97^\circ$ , which is $103^\circ$ to the nearest degree.