Find the values of x for which the vectors $\left( {\begin{array}{*{20}{c}}   1 \\   {2\cos x} \\   0 \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}}   { - 1} \\   {2\sin x} \\   1 \end{array}} \right)$ are perpendicular,  $0 \leqslant x \leqslant \frac{\pi }{2}$.

perpendicular when $\left( {\begin{array}{*{20}{c}}   1 \\   {2\cos x} \\   0 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}   { - 1} \\   {2\sin x} \\   1 \end{array}} \right) = 0$     (M1)

$\Rightarrow -1 + 4\sin x\cos x = 0$     A1

$\Rightarrow \sin 2x = \frac{1}{2}$     M1

$\Rightarrow 2x = \frac{\pi }{6},\frac{{5\pi }}{6}$

$\Rightarrow x = \frac{\pi }{{12}},\frac{{5\pi }}{{12}}$     A1A1 

Note: Accept answers in degrees. 

[5 marks]

Most candidates realised that the scalar product should be used to solve this problem and many obtained the equation $4\sin x\cos x = 1$. Candidates who failed to see that this could be written as $\sin 2x = 0.5$ usually made no further progress. The majority of those candidates who used this double angle formula carried on to obtain the solution $\frac{\pi }{{12}}$ but few candidates realised that $\frac{{5\pi }}{{12}}$ was also a solution.