A2 June 2020 Paper 1 Q11
11 The lines \(l_1\), \(l_2\) and \(l_3\) are defined as follows.
\[l_1 : \left(\mathbf{r} - \begin{bmatrix} 1 \\ 5 \\ -1 \end{bmatrix}\right) \times \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} = \mathbf{0}\]\[l_2 : \left(\mathbf{r} - \begin{bmatrix} -3 \\ 2 \\ 7 \end{bmatrix}\right) \times \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = \mathbf{0}\]\[l_3 : \left(\mathbf{r} - \begin{bmatrix} -5 \\ 12 \\ -4 \end{bmatrix}\right) \times \begin{bmatrix} 4 \\ 0 \\ 9 \end{bmatrix} = \mathbf{0}\](a)
(i) Explain how you know that two of the lines are parallel. [1 mark]
(ii) Show that the perpendicular distance between these two parallel lines is 7.95 units, correct to three significant figures. [5 marks]
(b) Show that the lines \(l_1\) and \(l_3\) meet, and find the coordinates of their point of intersection. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Explains why \(l_1\) and \(l_2\) are parallel, with reference to the direction vector of \(l_1\) being \(-1 \times\) the direction vector of \(l_2\). | E1 | 2.4 |
| (ii) Forms an expression for a vector from a point on \(l_1\) to a point on \(l_2\). PI by \(\sqrt{89}\) | M1 | 3.1a |
| (ii) Obtains a correct \(\overrightarrow{AB}\). PI by \(\sqrt{89}\) or Obtains a correct parametrised form for \(\overrightarrow{PQ}\) | A1 | 1.1b |
| (ii) Forms a vector product of \(\overrightarrow{AB}\) and the direction vector of \(l_1\) or \(l_2\) or Forms a scalar product of \(\overrightarrow{PQ}\) and the direction vector of \(l_1\) or \(l_2\) | M1 | 3.1a |
| (ii) Uses either distance \(= k\left|\overrightarrow{AB} \times \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\right|\) or \(\overrightarrow{PQ} \cdot \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = 0\) | M1 | 1.1a |
| (ii) Completes a rigorous argument to show that distance = 7.95 AG | R1 | 2.1 |
Typical solution
(i)
\[\begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} = -\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\]The direction vectors for \(l_1\) and \(l_2\) are multiples of each other, so \(l_1\) and \(l_2\) are parallel lines.
(ii)
Given a point \(A\) on \(l_1\) and a point \(B\) on \(l_2\), distance \(= |AB\sin\theta|\), where either line makes an angle \(\theta\) with \(AB\).
\[A(1, 5, -1),\ B(-3, 2, 7); \quad \overrightarrow{AB} = \begin{bmatrix} -4 \\ -3 \\ 8 \end{bmatrix}\]Unit direction vector \(\hat{l} = \dfrac{1}{\sqrt{14}}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\)
\[\overrightarrow{AB} \times \hat{l} = \frac{1}{\sqrt{14}}\begin{vmatrix} \mathbf{i} & -4 & 2 \\ \mathbf{j} & -3 & -1 \\ \mathbf{k} & 8 & 3 \end{vmatrix} = \frac{1}{\sqrt{14}}\begin{bmatrix} -1 \\ 28 \\ 10 \end{bmatrix}\]Required distance \(= \dfrac{1}{\sqrt{14}}\left(\sqrt{1^2 + 28^2 + 10^2}\right)\)
\(= 7.95\) (3 sig fig) as required.
Alternative typical solution:
\[\overrightarrow{PQ} = \begin{bmatrix} 4 - 2\lambda \\ 3 + \lambda \\ -8 - 3\lambda \end{bmatrix}\]\[0 = \begin{bmatrix} 4 - 2\lambda \\ 3 + \lambda \\ -8 - 3\lambda \end{bmatrix} \cdot \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix}\]\[0 = -8 + 4\lambda + 3 + \lambda + 24 + 9\lambda\]\[\lambda = \frac{-19}{14}\]\[\overrightarrow{PQ}_{min} = \begin{bmatrix} 47/7 \\ 23/14 \\ -55/14 \end{bmatrix}\]Required distance
\[= \sqrt{(47/7)^2 + (23/14)^2 + (55/14)^2}\]\(= 7.95\) (3 sig fig)
| Scheme | Marks | AO |
|---|---|---|
| Expresses the two lines in parametric form. Condone the same parameter for both lines. | M1 | 3.1a |
| Equates the two lines with two different parameters. | M1 | 1.1a |
| Solves two equations, from components of the vectors, simultaneously. | M1 | 1.1b |
| Verifies that the third components are equal. | A1 | 2.2a |
| Completes a rigorous argument by concluding that the third equation is satisfied and finding the coordinates of the point of intersection. | R1 | 2.1 |
| (11 marks) |
Typical solution
\[\mathbf{r} = \begin{bmatrix} 1 \\ 5 \\ -1 \end{bmatrix} + \mu\begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix}\]\[\mathbf{r} = \begin{bmatrix} -5 \\ 12 \\ -4 \end{bmatrix} + \lambda\begin{bmatrix} 4 \\ 0 \\ 9 \end{bmatrix}\]\[\begin{aligned} x &\quad 1 - 2\mu = -5 + 4\lambda \\ y &\quad 5 + \mu = 12 \\ z &\quad -1 - 3\mu = -4 + 9\lambda \end{aligned}\]Eqn. 2 \(\Rightarrow \mu = 7\)
Sub. in eqn. 1 giving \(\lambda = -2\)
\(\mu = 7\) and \(\lambda = -2\) satisfy eqn. 3, and the lines meet at \((-13, 12, -22)\)