AS October 2020 Paper 1 Q7
7 The equations of two intersecting lines are
\[\mathbf{r} = \begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \qquad \mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} + \mu\begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}\]where \(a\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{b} = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}\) | M1 | 1.1a |
| \(\mathbf{b} = \begin{pmatrix} -3 \\ -1 \\ 8 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
M1: Cross product either way round.
Allow inclusion of \(\lambda\) and/or \(\mu\) for M1 only.
A1: BC
or any non-zero numerical multiple.
Alternative method
| Scheme | Marks |
|---|---|
| \(\mathbf{b} = \begin{pmatrix} p \\ q \\ r \end{pmatrix}\) and \(\mathbf{b}.\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = 0\) and \(\mathbf{b}.\begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix} = 0\) | M1 |
| eg \(p = 1\) and \(2q + r = -2\) and \(q - r = 3\) leading to \(\mathbf{b} = \dfrac{1}{3}\begin{pmatrix} 3 \\ 1 \\ -8 \end{pmatrix}\) | A1 |
M1: \(p\), \(q\) or \(r\) could be any non-zero number
| Scheme | Marks | AO |
|---|---|---|
| Since lines intersect \(\begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} + \mu\begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}\) for some \(\lambda\) and \(\mu\) \(\therefore \begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix}.\mathbf{b} + \lambda\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}.\mathbf{b} = \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix}.\mathbf{b} + \mu\begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}.\mathbf{b}\) but | M1 | 1.1 |
| \(\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}.\mathbf{b} = \begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}.\mathbf{b} = 0 \Rightarrow \mathbf{b}.\begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} = \mathbf{b}.\begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
A1: AG
(corrected from the printed mark scheme: the last line printed the third component of \(\begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix}\) as 1; it is \(-1\), as in the question)
Alternative method
| Scheme | Marks |
|---|---|
| Find where the two lines meet and obtain \(a = -6\) (as below) \(\begin{pmatrix} -3 \\ -1 \\ 8 \end{pmatrix}.\begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} -3 \\ -1 \\ 8 \end{pmatrix}.\begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix}\) \(\Rightarrow 36 - a - 8\) and \(-6 + 40\) | M1 |
| \(36 - -6 - 8\) \(= 34\) \(= -6 + 40\) So dot products are equal | A1 |
This can be awarded 2 marks in part c as long as some comment is made in part c (such as “\(a=-6\)” or “see above”)
If using this method in part b then the intersection of the lines AND formulation of the dot products needs to happen for M1
M1: Correct formulation of both their dot products using their \(\mathbf{b}\)
A1: Conclusion needed
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} -3 \\ -1 \\ 8 \end{pmatrix}.\begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \\ 8 \end{pmatrix}.\begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} \Rightarrow 36 - a - 8 = -6 + 40\) | M1ft | 1.1a |
| \(a = -6\) | A1 ft | 1.1 |
| [2] |
Notes
M1ft: Correct formation of both dot products using their \(\mathbf{b}\)
Stating \(a=-6\) fine for 2 marks
Also allow “see above”
No marks if blank, even if \(a = -6\) seen in (b)
Alternative method
| Scheme | Marks |
|---|---|
| \(-12 + 2\lambda = 2 - 3\mu\) & \(-1 + \lambda = 5 - \mu\) \(\Rightarrow \lambda = 4,\ \mu = 2\) | M1ft |
| \(a = \mu - 2\lambda = -6\) | A1 ft |
M1ft: Forming \(x\) and \(z\) equations using their \(\mathbf{b}\) and solving for \(\lambda\) and \(\mu\)
A1 ft: From \(y\) equation