A2 June 2019 Paper 2 Q11

11 The line \(L_1\) has equation

\[\frac{x - 2}{3} = \frac{y + 4}{8} = \frac{4z - 5}{5}\]

The line \(L_2\) has equation

\[\left(\mathbf{r} - \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}\right) \times \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} = \mathbf{0}\]

Find the shortest distance between the two lines, giving your answer to three significant figures. [8 marks]

Mark scheme

Scheme	Marks	AO
Obtains a position vector of a point on \(L_1\).	B1	2.5
Obtains a direction vector for \(L_1\) ISW	B1	1.1b
Obtains a position vector of a point on \(L_2\) and a direction vector for \(L_2\)	B1	2.5
Obtains a correct vector between the two lines. Follow through from their position vectors	B1F	1.1b
Calculates the vector product of their two direction vectors or calculates the scalar product of the general vector between the lines with both direction vectors to obtain a pair of simultaneous equations	M1	3.1a
Obtains the correct vector product or obtains a correct pair of simultaneous equations	A1	1.1b
Uses their vector product or the solutions to their simultaneous equations to calculate the shortest distance between the two lines	M1	3.1a
Obtains the correct shortest distance. Allow awrt 5.18 Accept exact answer	A1	1.1b
	(8 marks)

Typical solution

\[\mathbf{r}_1 = \begin{pmatrix} 2 \\ -4 \\ 5/4 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ 8 \\ 5/4 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 5/4 \end{pmatrix} + \mu\begin{pmatrix} 12 \\ 32 \\ 5 \end{pmatrix}\]\[\mathbf{r}_2 = \begin{pmatrix} -2 \\ 0 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\]\[\begin{pmatrix} 2 \\ -4 \\ 5/4 \end{pmatrix} - \begin{pmatrix} -2 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \\ -7/4 \end{pmatrix}\]\[\begin{pmatrix} 12 \\ 32 \\ 5 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 91 \\ -26 \\ -52 \end{pmatrix}\]\[\frac{\begin{pmatrix} 4 \\ -4 \\ -\frac{7}{4} \end{pmatrix}.\begin{pmatrix} 91 \\ -26 \\ -52 \end{pmatrix}}{13\sqrt{69}} = \frac{43}{\sqrt{69}}\]\[= 5.17659\ldots = 5.18 \text{ 3sf}\]