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 $\left| {\overrightarrow {{\text{OP}}} } \right|$.