Find a vector equation of the line that passes through P and Q.

The line through P and Q is perpendicular to the vector 2i $+$ nk. Find the value of $n$.

valid attempt to find direction vector     (M1)

eg$\,\,\,\,\,$$\overrightarrow {{\text{PQ}}} ,{\text{ }}\overrightarrow {{\text{QP}}}$

correct direction vector (or multiple of)     (A1)

eg$\,\,\,\,\,$6i $+$ j $-$ 3k

any correct equation in the form r $=$ a $+$ tb (any parameter for $t$)     A2     N3

where a is i $+$ 2j $-$ k or 7i $+$ 3j $-$ 4k , and b is a scalar multiple of 6i $+$ j $-$ 3k

eg$\,\,\,\,\,$r $=$ 7i $+$ 3j $-$ 4k $+$ t(6i $+$ j $-$ 3k), r $= \left( {\begin{array}{*{20}{c}} {1 + 6s} \\ {2 + 1s} \\ { - 1 - 3s} \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 6} \\ { - 1} \\ 3 \end{array}} \right)$

Notes: Award A1 for the form a $+$ tb, A1 for the form L $=$ a $+$ tb, A0 for the form r $=$ b $+$ ta.

[4 marks]

correct expression for scalar product     (A1)

eg$\,\,\,\,\,$$6 \times 2 + 1 \times 0 + ( - 3) \times n,{\text{ }} - 3n + 12$

setting scalar product equal to zero (seen anywhere)     (M1)

eg$\,\,\,\,\,$u $\bullet$ v $= 0,{\text{ }} - 3n + 12 = 0$

$n = 4$    A1     N2

[3 marks]