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.