Find $\overrightarrow {{\rm{AB}}}$ .

Find an equation for ${L_1}$ in the form ${\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}$ .

Find the angle between ${L_1}$ and ${L_2}$ .

The lines ${L_1}$ and ${L_2}$ intersect at point C. Find the coordinates of C.

appropriate approach     (M1)

e.g. $\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}}$ , $B - A$

$\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right)$     A1     N2

[2 marks]

any correct equation in the form ${\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}$     A2 N2

where ${\boldsymbol{b}}$ is a scalar multiple of $\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right)$

e.g. ${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 4 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right)$ , ${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} {2 + t}\\ { - 2 - t}\\ {5 + t} \end{array}} \right)$ , ${\boldsymbol{r}} = 2{\boldsymbol{i}} - 2{\boldsymbol{j}} + 5{\boldsymbol{k}} + t({\boldsymbol{i}} - {\boldsymbol{j}} + {\boldsymbol{k}})$

 [2 marks]

choosing correct direction vectors $\left( \begin{array}{l} 2\\ 1\\ 3 \end{array} \right)$ , $\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right)$     (A1)(A1)

finding scalar product and magnitudes     (A1)(A1)(A1)

scalar product $= 1 \times 2 + - 1 \times 1 + 1 \times 3$ $\left( { = 4} \right)$

magnitudes $\sqrt {{1^2} + {{( - 1)}^2} + {1^2}}{\text{  }}( = 1.73 \ldots )$ , $\sqrt {4 + 1 + 9}{\text{  }}( = 3.74 \ldots )$

substitution into $\frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}$ (accept $\theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}$ , but not $\sin \theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}$ )     M1

e.g. $\cos \theta = \frac{{1 \times 2 + - 1 \times 1 + 1 \times 3}}{{\sqrt {{1^2} + {{( - 1)}^2} + {1^2}} \sqrt {{2^2} + {1^2} + {3^2}} }}$ , $\cos \theta = \frac{4}{{\sqrt {42} }}$

$\theta = 0.906$ $({51.9^ \circ })$     A1      N5

[7 marks]

METHOD 1 (from ${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}   1 \\   { - 1} \\   4 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}}   1 \\   { - 1} \\   1 \end{array}} \right)$ )

appropriate approach     (M1)

e.g. ${\boldsymbol{p}} = {\boldsymbol{r}}$ , $\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 4 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right)t = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ 7 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2\\ 1\\ 3 \end{array}} \right)s$

two correct equations    A1A1

e.g. $1 + t = 2 + 2s$ , $- 1 - t = 4 + s$ , $4 + t = 7 + 3s$

attempt to solve     (M1)

one correct parameter     A1

e.g. $t = - 3$ , $s = - 2$

C is $( - 2{\text{, }}2{\text{, }}1)$     A1     N3

METHOD 2 (from ${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}   2 \\   { - 2} \\   5 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}}   1 \\   { - 1} \\   1 \end{array}} \right)$ )

appropriate approach     (M1)

e.g. ${\boldsymbol{p}} = {\boldsymbol{r}}$ , $\left( {\begin{array}{*{20}{c}} 2\\ { - 2}\\ 5 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ 7 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2\\ 1\\ 3 \end{array}} \right)s$

two correct equations     A1A1

e.g. $2 + t = 2 + 2s$ , $- 2 - t = 4 + s$ , $5 + t = 7 + 3s$

attempt to solve     (M1)

one correct parameter     A1

e.g. $t = - 4$ , $s = - 2$

C is $( - 2{\text{, }}2{\text{, }}1)$     A1     N3

[6 marks]

Finding $\overrightarrow {{\rm{AB}}}$ was generally well done, although some candidates reversed the subtraction.

In part (b) not all the candidates recognized that $\overrightarrow {{\rm{AB}}}$ was the direction vector of the line, as some used the position vector of point B as the direction vector.

Many candidates successfully used scalar product and magnitudes in part (c), although a large number did choose vectors other than the direction vectors and many did not state clearly which vectors they were using.

Candidates who were comfortable on the first three parts often had little difficulty with the final part. While the resulting systems were easily solved algebraically, a surprising number of candidates did not check their solutions either manually or with technology. An occasionally seen error in the final part was using a midpoint to find C. Some candidates found the point of intersection in part (c) rather than in part (d), indicating a familiarity with the type of question but a lack of understanding of the concepts involved.