Find

(i)     $\overrightarrow {{\rm{AB}}}$ ;

(ii)    $\overrightarrow {{\rm{AD}}}$ giving your answer in terms of k .

[3 marks]

Show that $k = 7$ .

The point C is such that $\overrightarrow {{\rm{BC}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}}$ .

Find the position vector of C.

Find $\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}}$ .

(i) evidence of combining vectors     (M1)

e.g. $\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}}$  (or $\overrightarrow {{\rm{AD}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}}$ in part (ii)) 

$\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 2\\ { - 4}\\ { - 2} \end{array}} \right)$    A1     N2

(ii) $\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}} 2\\ {k - 5}\\ { - 2} \end{array}} \right)$     A1     N1

[3 marks]

evidence of using perpendicularity $\Rightarrow$ scalar product = 0     (M1)

e.g. $\left( {\begin{array}{*{20}{c}} 2\\ { - 4}\\ { - 2} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2\\ {k - 5}\\ { - 2} \end{array}} \right) = 0$

$4 - 4(k - 5) + 4 = 0$     A1

$- 4k + 28 = 0$ (accept any correct equation clearly leading to $k = 7$ )    A1

$k = 7$     AG     N0

[3 marks]

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

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

evidence of correct approach     (M1)

e.g. $\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{BC}}}$ , $\left( {\begin{array}{*{20}{c}} 3\\ 1\\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right)$ , $\left( {\begin{array}{*{20}{c}} {x - 3}\\ {y - 1}\\ {z - 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right)$

$\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}} 4\\ 2\\ 1 \end{array}} \right)$     A1     N3

[4 marks]

METHOD 1

choosing appropriate vectors, $\overrightarrow {{\rm{BA}}}$ , $\overrightarrow {{\rm{BC}}}$     (A1)

finding the scalar product     M1

e.g. $- 2(1) + 4(1) + 2( - 1)$ , $2(1) + ( - 4)(1) + ( - 2)( - 1)$

$\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}$     A1     N1

METHOD 2

$\overrightarrow {{\rm{BC}}}$ parallel to $\overrightarrow {{\rm{AD}}}$ (may show this on a diagram with points labelled)     R1

$\overrightarrow {{\rm{BC}}} \bot \overrightarrow {{\rm{AB}}}$ (may show this on a diagram with points labelled)     R1

${\rm{A}}\widehat {\rm{B}}{\rm{C}} = 90^\circ$

$\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}$     A1     N1

[3 marks]

This question was well done by many candidates. Most found $\overrightarrow {{\rm{AB}}}$ and $\overrightarrow {{\rm{AD}}}$ correctly.

The majority of candidates correctly used the scalar product to show $k = 7$ .

Some confusion arose in substituting $k = 7$ into $\overrightarrow {{\rm{AD}}}$ , but otherwise part (c) was well done, though finding the position vector of C presented greater difficulty.

Owing to $\overrightarrow {{\rm{AB}}}$ and $\overrightarrow {{\rm{BC}}}$ being perpendicular, no problems were created by using these two vectors to find $\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = 0$ , and the majority of candidates answering part (d) did exactly that.