Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
2\\
{ - 3}\\
6
\end{array}} \right)\) and  \({\boldsymbol{w}} = \left( {\begin{array}{*{20}{c}}
k\\
{ - 2}\\
4
\end{array}} \right)\) , for \(k > 0\) . The angle between v and w is \(\frac{\pi }{3}\) .

Find the value of \(k\) .

correct substitutions for \({\boldsymbol{v}} \bullet {\boldsymbol{w}}\) ; \(\left| {\boldsymbol{v}} \right|\) ; \(\left| {\boldsymbol{w}} \right|\)     (A1)(A1)(A1)

e.g. \(2k + ( - 3) \times ( - 2) + 6 \times 4\) , \(2k + 30\) ; \(\sqrt {{2^2} + {{( - 3)}^2} + {6^2}} \) , \(\sqrt {49} \) ; \(\sqrt {{k^2} + {{( - 2)}^2} + {4^2}} \) , \(\sqrt {{k^2} + 20} \)

evidence of substituting into the formula for scalar product     (M1)

e.g. \(\frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)

correct substitution     A1

e.g. \(\cos \frac{\pi }{3} = \frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)

\(k = 18.8\)     A2     N5

[7 marks]

For the most part, this question was well done and candidates had little difficulty finding the scalar product, the appropriate magnitudes and then correctly substituting into the formula for the angle between vectors. However, few candidates were able to solve the resulting equation using their GDCs to obtain the correct answer. Problems arose when candidates attempted to solve the resulting equation analytically.