Question
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, \(0 \leqslant x \leqslant \frac{\pi }{2}\).
▶️Answer/Explanation
Markscheme
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]
Examiners report
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.