WAEC 2011 · Paper 2 · Q16

(a)
The position vectors of points $P$ , $Q$ and $R$ are $5\mathbf i + 3\mathbf j$ , $8\mathbf i - \mathbf j$ and $11\mathbf i - 5\mathbf j$ respectively. (i) Show that $P$ , $Q$ and $R$ are collinear. (ii) Find the scalars $k_1$ and $k_2$ such that $37\mathbf i - \mathbf j = k_1\mathbf p + k_2\mathbf r$ where $\mathbf p$ and $\mathbf r$ are the position vectors of $P$ and $R$ respectively.
Separate values with commas, e.g. 3, −2
(b)
Given that $\mathbf m = 3\mathbf i - 4\mathbf j$ and $\mathbf n = 6\mathbf i + 4\mathbf j$ , find the angle between the two vectors, correct to the nearest degree.

Worked solution (try it first)

$\overrightarrow{PQ} = (8 - 5)\mathbf i + (-1 - 3)\mathbf j$
$= 3\mathbf i - 4\mathbf j$ .
$\overrightarrow{QR} = (11 - 8)\mathbf i + (-5 + 1)\mathbf j$
$= 3\mathbf i - 4\mathbf j$ .
They are equal, so parallel, and they share $Q$ : $P$ , $Q$ and $R$ are collinear.

$\mathbf m \cdot \mathbf n = 18 - 16 = 2$ , $|\mathbf m| = 5$ and $|\mathbf n| = \sqrt{52}$ .
$\cos\theta = \dfrac{2}{5\sqrt{52}}$
$\approx 0.0555$ , so $\theta \approx 87^\circ$ .