WAEC 2012 · Paper 2 · Q5

Vectors

(a)
Two sides of a triangle are represented by the vectors $\mathbf p = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf q = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ . (i) Show that they are perpendicular to each other. (ii) Find the area of the triangle. (iii) Find the angles of the triangle.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(i)

$\mathbf p \cdot \mathbf q = (3)(2) + (-2)(3)$
$= 6 - 6$
$= 0$ , so they are perpendicular.

(ii)

$|\mathbf p| = \sqrt{9 + 4} = \sqrt{13}$ and $|\mathbf q| = \sqrt{4 + 9} = \sqrt{13}$ .
The right angle is between them, so they are the base and height: area $= \frac12 \times \sqrt{13} \times \sqrt{13}$
$= 6.5$ square units.

(iii)

One angle is $90^\circ$ .
The two sides are equal, so the other angles are equal: $\frac{180^\circ - 90^\circ}{2} = 45^\circ$ each.

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