WAEC 2011 · Paper 2 · Q17

(a)
$m\begin{pmatrix} 2 \\ 1 \end{pmatrix} + n\begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}$ where $m$ and $n$ are scalars. Find the value of $(m + n)$ .
(b)
$A(-1, 3)$ , $B(2, -1)$ and $C(5, 3)$ are the vertices of $\triangle ABC$ . (i) Express in column notation the unit vectors parallel to $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . (ii) Use a dot product to calculate $B\hat AC$ , correct to the nearest degree.

Worked solution (try it first)

$\overrightarrow{AB} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ , of length 5.
$\overrightarrow{AC} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}$ , of length 6.
Unit vectors: $\frac15\begin{pmatrix} 3 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ .

$\overrightarrow{AB} \cdot \overrightarrow{AC} = 18 + 0$
$= 18$ .
$\cos B\hat AC = \dfrac{18}{5 \times 6}$
$= 0.6$ , so $B\hat AC \approx 53^\circ$ .