Write down the equation of ${L_1}$ in the form ${\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}$.

The line ${L_2}$ has equation ${\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right)$ . The point A has position vector $\left( {\begin{array}{*{20}{c}} 6\\ 2\\ 9 \end{array}} \right)$ .

Show that A lies on ${L_2}$ .

Let B be the point of intersection of lines ${L_1}$ and ${L_2}$ .

(i)     Show that $\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ {14} \end{array}} \right)$ .

(ii)    Find $\overrightarrow {{\rm{AB}}}$ .

The point C is at (2, 1, − 4). Let D be the point such that ABCD is a parallelogram.

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

${L_1}:{\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ 0 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right)$     A2     N2

[2 marks]

evidence of equating ${\boldsymbol{r}}$ and $\overrightarrow {{\rm{OA}}}$     (M1)

e.g. $\left( {\begin{array}{*{20}{c}} 6\\ 2\\ 9 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right)$ , $A = r$

one correct equation     A1

e.g. $6 = 2 + 2s$ , $2 = 4 - s$ , $9 = - 1 + 5s$

$s = 2$     A1

evidence of confirming for other two equations     A1

e.g. $6 = 2 + 4$ , $2 = 4 - 2$ , $9 = - 1 + 10$

so A lies on ${L_2}$      AG     N0

[4 marks]

(i) evidence of approach     M1

e.g. $\left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ 0 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right)$ ${L_1} = {L_2}$

one correct equation     A1

e.g. $2 + 2s = 8$ , $4 - s = 1$ , $- 1 + 5s = t$

attempt to solve     (M1)

finding $s = 3$     A1

substituting     M1

e.g. $\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}} \right) + 3\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right)$

$\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ {14} \end{array}} \right)$     AG     N0

(ii) evidence of appropriate approach     (M1)

e.g. $\overrightarrow {{\rm{AB}}}  = \overrightarrow {{\rm{AO}}}  + \overrightarrow {{\rm{OB}}}$ , $\overrightarrow {{\rm{AB}}}  = \overrightarrow {{\rm{OB}}}  - \overrightarrow {{\rm{OA}}}$

$\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right)$     A1     N2  

[7 marks]

evidence of appropriate approach    (M1)

e.g. $\overrightarrow {{\rm{AB}}}  = \overrightarrow {{\rm{DC}}}$

correct values     A1

e.g. $\overrightarrow {{\rm{OD}}}  + \left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ 1\\ { - 4} \end{array}} \right)$ ,  $\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ 1\\ { - 4} \end{array}} \right)$ , $\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2 - x}\\ {1 - y}\\ { - 4 - z} \end{array}} \right)$

$\overrightarrow {{\rm{OD}}}  = \left( {\begin{array}{*{20}{c}} 0\\ 2\\ { - 9} \end{array}} \right)$     A1     N2

[3 marks]

Very few candidates gave a correct direction vector parallel to the z-axis. Provided they wrote down an equation here they were able to earn most of subsequent marks on follow through.

For (b), many found the correct parameter but neglected to confirm it in the other two equations.

In (c) some performed a trial and error approach to obtaining an integer parameter and thus did not "show" the mathematical origin of the result. Finding vector $\overrightarrow {{\rm{AB}}}$ proved accessible.

A good number of candidates had an appropriate approach to (d), although surprisingly many subtracted $\overrightarrow {{\rm{OC}}}$ from $\overrightarrow {{\rm{AB}}}$ in finding $\overrightarrow {{\rm{OD}}}$ .