(a)     Find the coordinates of the point \(A\) on \({l_1}\) and the point \(B\) on \({l_2}\) such that \(\overrightarrow {{\text{AB}}} \) is perpendicular to both \({l_1}\) and \({l_2}\) .

(b)     Find \(\left| {{\text{AB}}} \right|\) .

(c)     Find the Cartesian equation of the plane \(\prod \) which contains \({l_1}\) and does not intersect \({l_2}\) .

(a)     on \({l_1}\)   A(\( - 3 + 3\lambda \), \( - 4 + 2\lambda \), \(6 - 2\lambda \))     A1

on \({l_2}\)   \({l_2}:r = \left( {\begin{array}{*{20}{c}}
  4 \\
  { - 7} \\
  3
\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  4 \\
  { - 1}
\end{array}} \right)\)     (M1)

\( \Rightarrow \) B(\(4 - 3\mu \), \( - 7 + 4\mu \), \( - 3 - \mu \))     A1

\(\overrightarrow {{\text{BA}}}  = {\boldsymbol{a}} - {\boldsymbol{b}} = \left( {\begin{array}{*{20}{c}}
  {3\lambda  + 3\mu  - 7} \\
  {2\lambda  - 4\mu  + 3} \\
  { - 2\lambda  + \mu  + 9}
\end{array}} \right)\)     (M1)A1

EITHER

\({\text{BA}} \bot {l_1} \Rightarrow {\text{BA}} \cdot \left( {\begin{array}{*{20}{c}}
  3 \\
  2 \\
  { - 2}
\end{array}} \right) = 0 \Rightarrow 3\left( {3\lambda  + 3\mu  - 7} \right) + 2\left( {2\lambda  - 4\mu  + 3} \right) - 2\left( { - 2\lambda  + \mu  + 9} \right) = 0\)     M1

\( \Rightarrow 17\lambda  - \mu  = 33\)     A1

\({\text{BA}} \bot {l_2} \Rightarrow {\text{BA}} \cdot \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  4 \\
  { - 1}
\end{array}} \right) = 0 \Rightarrow  - 3\left( {3\lambda  + 3\mu  - 7} \right) + 4\left( {2\lambda  - 4\mu  + 3} \right) - \left( { - 2\lambda  + \mu  + 9} \right) = 0\)     M1

\( \Rightarrow \lambda  - 26\mu  = - 24\)     A1

solving both equations above simultaneously gives
\(\lambda  = 2\); \(\mu  = 1 \Rightarrow \) A(3, 0, 2), B(1, –3, –4)     A1A1A1A1

OR

\(\left| {\begin{array}{*{20}{c}}
  {\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
  3&2&{ - 2} \\
  { - 3}&4&{ - 1}
\end{array}} \right| = 6{\boldsymbol{i}} + 9{\boldsymbol{j}} + 18{\boldsymbol{k}}\)    M1A1

so \(\overrightarrow {{\text{AB}}}  = p\left( {\begin{array}{*{20}{c}}
  2 \\
  3 \\
  6
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {3\lambda  + 3\mu  - 7} \\
  {2\lambda  - 4\mu  + 3} \\
  { - 2\lambda  + \mu  + 9}
\end{array}} \right)\)     M1A1

\({3\lambda  + 3\mu  - 2p = 7}\)
\({2\lambda  - 4\mu  - 3p = - 3}\)
\({ - 2\lambda  + \mu  - 6p = - 9}\)

\(\lambda  = 2\), \(\mu  = 1\), \(p = 1\)     A1A1

A(\( - 3 + 6\), \( - 4 + 4\), \(6 - 4\)) \(=\) (\(3\), \(0\), \(2\))     A1

B(\(4 - 3\), \( - 7 + 4\), \( - 3 - 1\)) \(=\) (\(1\), \( - 3\), \( - 4\))     A1

[13 marks]

(b)     \({\text{AB}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 3} \\
  { - 4}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  3 \\
  0 \\
  2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 2} \\
  { - 3} \\
  { - 6}
\end{array}} \right)\)     (A1)

\(\left| {{\text{AB}}} \right| = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( { - 6} \right)}^2}}  = \sqrt {49}  = 7\)     M1A1

[3 marks]

(c)     from (b) \(2{\boldsymbol{i}} + 3{\boldsymbol{j}} + 6{\boldsymbol{k}}\) is normal to both lines

\({l_1}\) goes through (–3, –4, 6) \( \Rightarrow \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  { - 4} \\
  6
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  2 \\
  3 \\
  6
\end{array}} \right) = 18\)     M1A1

hence, the Cartesian equation of the plane through \({l_1}\) , but not \({l_2}\) , is \(2x + 3y + 6z = 18\)     A1

[3 marks]

Total [19 marks]

There were a lot of arithmetic errors in the treatment of this question, even though it was apparent that many students did understand the methods involved. In (a) many students failed to realise that \(\overrightarrow {{\text{AB}}} \) should be a multiple of the cross product of the two direction vectors, rather than the cross product itself, and many students failed to give the final answer as coordinates.