Find \(\overrightarrow {{\text{AB}}} \).

Find the coordinates of C.

Write down a vector equation for \(L\).

Given that \(\left| {\overrightarrow {{\text{AB}}} } \right| = k\left| {\overrightarrow {{\text{AC}}} } \right|\), find \(k\).

The point D lies on \(L\) such that \(\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{BD}}} } \right|\). Find the possible coordinates of D.

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{B}} - {\text{A}},{\text{ AO}} + {\text{OB}},{\text{ }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1 \\ 5 \\ { - 7} \end{array}} \right)\)

\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { - 10} \\ 4 \\ 1 \end{array}} \right)\)     A1     N2

[2 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{OC}} = {\text{OA}} + {\text{AC}},{\text{ }}\left( {\begin{array}{*{20}{c}} {1 + 6} \\ {5 - 4} \\ { - 7 + 0} \end{array}} \right)\)

\({\text{C}}(7,{\text{ }}1,{\text{ }} - 7)\)    A1     N2

[2 marks]

any correct equation in the form r \( = \) a \( + t\)b (accept any parameter for \(t\)

where a is \(\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right)\), and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)     A2     N2

eg\(\,\,\,\,\,\)r \( = \left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\), r \( =  - 9\)i \( + 9\)j \( - 6\)k \( + s(6\)i \( - 4\)j \( + 0\)k\()\)

[2 marks]

correct magnitudes     (A1)(A1)

eg\(\,\,\,\,\,\)\(\sqrt {{{( - 10)}^2} + {{( - 4)}^2} + {1^2}} ,{\text{ }}\sqrt {{6^2} + {{( - 4)}^2} + {{(0)}^2}} ,{\text{ }}\sqrt {{{10}^2} + {4^2} + 1} ,{\text{ }}\sqrt {{6^2} + {4^2}} \)

\(k = \frac{{\sqrt {117} }}{{\sqrt {52} }}{\text{ }}( = 1.5){\text{ }}({\text{exact}})\)    A1     N3

[3 marks]

correct interpretation of relationship between magnitudes     (A1)

eg\(\,\,\,\,\,\)\({\text{AB}} = 1.5{\text{AC}},{\text{ BD}} = 1.5{\text{AC}},{\text{ }}\sqrt {117}  = \sqrt {52{t^2}} \)

recognizing D can have two positions (may be seen in working)     R1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{BD}}}  = 1.5\overrightarrow {{\text{AC}}} \) and \(\overrightarrow {{\text{BD}}}  =  - 1.5\overrightarrow {{\text{AC}}} ,{\text{ }}t =  \pm 1.5\), diagram, two answers

valid approach (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \overrightarrow {{\text{BD}}} {\text{, }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BD}}} = k\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)

one correct expression for \(\overrightarrow {{\text{OD}}} \)     (A1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + 1.5\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) - 1.5\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)

\({\text{D}} = (0,{\text{ }}3,{\text{ }} - 6),{\text{ D}} = ( - 18,{\text{ }}15,{\text{ }} - 6)\) (accept position vectors)     A1A1     N3

[6 marks]

Parts (a) and (b) were attempted by the great majority of the candidates and appropriate approaches were seen, earning at least the method marks.

Parts (a) and (b) were attempted by the great majority of the candidates and appropriate approaches were seen, earning at least the method marks.

Part (c) was generally well done, with many candidates writing the equation of the line as \(L = \) a \( + t\)b, losing one mark.

Part (d) was also well answered by a great majority of students. Even those candidates who had part (a) incorrect, could gain all the marks here.

Part (e) was the most challenging of the paper. For many a major problem was to set up the equation \(\sqrt {117}  = \sqrt {52{t^2}} \) and, hence, realize that D could have two positions.