WAEC 2014 · Paper 2 · Q14

(a)
Given that $\mathbf x = 3\mathbf i - \mathbf j$ , $\mathbf y = 2\mathbf i + k\mathbf j$ and the cosine of the angle between $\mathbf x$ and $\mathbf y$ is $\frac{\sqrt5}{5}$ , find the values of the constant $k$ .
Separate values with commas, e.g. 3, −2
(b)
In the quadrilateral $ABCD$ , $\overrightarrow{AB} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}$ , $\overrightarrow{AC} = \begin{pmatrix} -6 \\ -9 \end{pmatrix}$ and $\overrightarrow{BD} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}$ . Show whether or not $ABCD$ is a parallelogram.
Show the answer
$ABCD$ is a parallelogram

Worked solution (try it first)

$\mathbf x \cdot \mathbf y = 3(2) + (-1)(k)$
$= 6 - k$ , $|\mathbf x| = \sqrt{10}$ and $|\mathbf y| = \sqrt{4 + k^2}$ .
So $\dfrac{6 - k}{\sqrt{10}\sqrt{4 + k^2}} = \dfrac{\sqrt5}{5}$ , which gives $5(6 - k) = \sqrt{50}\sqrt{4 + k^2}$ .
Square both sides: $900 - 300k + 25k^2 = 200 + 50k^2$ .
Rearrange: $25k^2 + 300k - 700 = 0$ , so $k^2 + 12k - 28 = 0$ .
Factorise: $(k - 2)(k + 14) = 0$ , so $k = 2$ or $k = -14$ .
Both make $6 - k$ positive (4 and 20), so both satisfy the original equation.

$\overrightarrow{BC}$ is $\overrightarrow{AC} - \overrightarrow{AB}$ , which is $\begin{pmatrix} -1 \\ -8 \end{pmatrix}$ .
$\overrightarrow{AD}$ is $\overrightarrow{AB} + \overrightarrow{BD}$ , which is $\begin{pmatrix} -1 \\ -8 \end{pmatrix}$ .
So $\overrightarrow{AD} = \overrightarrow{BC}$ : $AD$ and $BC$ are equal and parallel.
Also $\overrightarrow{DC}$ is $\overrightarrow{AC} - \overrightarrow{AD}$ , which is $\begin{pmatrix} -5 \\ -1 \end{pmatrix}$ , the same as $\overrightarrow{AB}$ .
Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.