Find the angle between the lines $\frac{{x - 1}}{2} = 1 - y = 2z$ and $x = y = 3z$ .

consider a vector parallel to each line,

e.g. ${\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}   4 \\   { - 2} \\   1 \end{array}} \right)$ and ${\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}   3 \\   3 \\   1 \end{array}} \right)$     A1A1

let $\theta$ be the angle between the lines

$\cos \theta  = \frac{{\left| {{\boldsymbol{u \times v}}} \right|}}{{\left| {\boldsymbol{u}} \right|\left| {\boldsymbol{v}} \right|}} = \frac{{\left| {12 - 6 + 1} \right|}}{{\sqrt {21} \sqrt {19} }}$     M1A1

$= \frac{7}{{\sqrt {21} \sqrt {19} }} = 0.350...$     (A1)

so $\theta  = 69.5$ $\left( {{\text{or }}1.21{\text{ rad or }}\arccos \left( {\frac{7}{{\sqrt {21} \sqrt {19} }}} \right)} \right)$     A1     N4

Note: Allow FT from incorrect reasonable vectors.

[6 marks]

Most students knew how to find the angle between two vectors, although many could not find the correct two direction vectors.