A line \({L_1}\) has equation r = \(\left( \begin{array}{c} - 5\\ - 3\\2\end{array} \right) + \lambda \left( \begin{array}{c} - 1\\2\\2\end{array} \right)\).

A line \({L_2}\) passing through the origin intersects \({L_1}\) and is perpendicular to \({L_1}\).

(a)     Find a vector equation of \({L_2}\).

(b)     Determine the shortest distance from the origin to \({L_1}\).

(a)     METHOD 1

for P on \({L_1},{\text{ }}\overrightarrow {{\text{OP}}}  = \) \(\left( \begin{array}{c} - 5 - \lambda \\ - 3 + 2\lambda \\2 + 2\lambda \end{array} \right)\)

require \(\left( \begin{array}{c} - 5 - \lambda \\ - 3 + 2\lambda \\2 + 2\lambda \end{array} \right) \bullet \left( \begin{array}{c} - 1\\2\\2\end{array} \right) = 0\)     M1

\(5 + \lambda  - 6 + 4\lambda  + 4 + 4\lambda  = 0\) (or equivalent)     A1

\(\lambda  =  - \frac{1}{3}\)     A1

\(\therefore \overrightarrow {{\rm{OP}}}  = \left( \begin{array}{c} - \frac{{14}}{3}\\ - \frac{{11}}{3}\\\frac{4}{3}\end{array} \right)\)     A1

\({L_2}:\boldsymbol{r} = \mu \left( \begin{array}{c} - 14\\ - 11\\4\end{array} \right)\)     A1

 

Note:     Do not award the final A1 if \(\boldsymbol{r} =\) is not seen.

 

[5 marks]

METHOD 2

Calculating either \(\left| {\overrightarrow {{\text{OP}}} } \right|\) or \({\left| {\overrightarrow {{\text{OP}}} } \right|^2}\) eg

\(\left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt {{{( - 5 - \lambda )}^2} + {{( - 3 + 2\lambda )}^2} + {{(2 + 2\lambda )}^2}} \)     A1

\( = \sqrt {9{\lambda ^2} + 6\lambda  + 38} \)

Solving either \(\frac{{\text{d}}}{{{\text{d}}\lambda }}\left( {\left| {\overrightarrow {{\text{OP}}} } \right|} \right) = 0\) or \(\frac{{\text{d}}}{{{\text{d}}\lambda }}\left( {{{\left| {\overrightarrow {{\text{OP}}} } \right|}^2}} \right) = 0\) for \(\lambda \).     M1

\(\lambda  =  - \frac{1}{3}\)     A1

\(\overrightarrow {{\rm{OP}}}  = \left( \begin{array}{c} - \frac{{14}}{3}\\ - \frac{{11}}{3}\\\frac{4}{3}\end{array} \right)\)     A1

\({L_2}:\boldsymbol{r} = \mu \left( \begin{array}{c} - 14\\ - 11\\4\end{array} \right)\)      A1

 

Note:     Do not award the final A1 if \(\boldsymbol{r} =\) is not seen.

 

[5 marks]

 

(b)     METHOD 1

\(\left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt {{{\left( { - \frac{{14}}{3}} \right)}^2} + {{\left( { - \frac{{11}}{3}} \right)}^2} + {{\left( {\frac{4}{3}} \right)}^2}} \)     (M1)

\( = 6.08{\text{ }}\left( { = \sqrt {37} } \right)\)     A1

METHOD 2

shortest distance = \(\frac{{\left| {\left( \begin{array}{c} - 1\\2\\2\end{array} \right) \times \left( \begin{array}{c} - 5\\ - 3\\2\end{array} \right)} \right|}}{{\left| {\left( \begin{array}{c} - 1\\2\\2\end{array} \right)} \right|}}\)     (M1)

\( = \frac{{\left| {10i + 8j + 13k} \right|}}{{\sqrt {1 + 4 + 4} }}\)

\( = 6.08{\text{ }}\left( { = \sqrt {37} } \right)\)     A1

[2 marks]

 

Total [7 marks]

Part (a) was not well done. Most candidates recognised the need to calculate a scalar product. Some candidates made careless sign or arithmetic errors when solving for \(\lambda \). A few candidates neglected to express their final answer in the form ‘r  =’.

Candidates who were successful in answering part (a) generally answered part (b) correctly. The large majority of successful candidates calculated \(\left| {\overrightarrow {{\text{OP}}} } \right|\).