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.