Let u \( = \left( {\begin{array}{*{20}{c}}
2\\
3\\
{ - 1}
\end{array}} \right)\) and w \( = \left( {\begin{array}{*{20}{c}}
3\\
{ - 1}\\
p
\end{array}} \right)\) . Given that u is perpendicular to w , find the value of p .

Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  q \\
  5
\end{array}} \right)\) . Given that \(\left| {\boldsymbol{v}} \right| = \sqrt {42} \), find the possible values of \(q\) .

evidence of equating scalar product to 0     (M1)

\(2 \times 3 + 3 \times ( - 1) + ( - 1) \times p = 0\)     (\(6 - 3 - p = 0\), \(3 - p = 0\))     A1

\(p = 3\)     A1     N2

[3 marks]

evidence of substituting into magnitude formula     (M1)

e.g. \(\sqrt {1 + {q^2} + 25} \) , \(1 + {q^2} + 25\)

setting up a correct equation     A1

e.g. \(\sqrt {1 + {q^2} + 25}  = \sqrt {42} \) , \(1 + {q^2} + 25 = 42\) , \({q^2} = 16\)

\(q = \pm 4\)     A1    N2

[3 marks]

Most candidates knew to set the scalar product equal to zero.

Most candidates knew to set the scalar product equal to zero. A pleasing number found both answers for \(q\), although some often neglected to provide both solutions.