AS June 2023 Paper 1 Q2
2 The lines \(L_1\) and \(L_2\) have the following equations.
\(L_1 : \mathbf{r} = \begin{pmatrix} -5 \\ 6 \\ 15 \end{pmatrix} + \lambda\begin{pmatrix} 5 \\ -2 \\ -2 \end{pmatrix}\)
\(L_2 : \mathbf{r} = \begin{pmatrix} 24 \\ 1 \\ -5 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ 1 \\ -4 \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| \(-5 + 5\lambda = 24 + 3\mu\) & \(6 - 2\lambda = 1 + \mu\) | B1 | 1.1 |
| \(5\lambda - 3\mu = 29\) & \(6\lambda + 3\mu = 15\) | M1 | 1.1 |
| \(\lambda = 4\) & \(\mu = -3\) | A1 | 1.1 |
| \(\lambda = 4\) & \(\mu = -3 \Rightarrow \text{LHS} = 15 - 2 \times 4 = 7\) and \(\text{RHS} = -5 - 4 \times -3 = 7 = \text{LHS}\) so all 3 equations are satisfied so \(L_1\) and \(L_2\) do intersect | A1 | 1.1 |
| \(\begin{pmatrix} 15 \\ -2 \\ 7 \end{pmatrix}\) | A1 | 1.1 |
| [5] |
Notes
B1: Forming 2 correct equations in \(\lambda\) and \(\mu\).
Third equation is \(15 - 2\lambda = -5 - 4\mu\)
M1: Attempt to solve (eg scaling one equation and adding; or rewriting to a standard form for solution BC)
If scaling or substituting method must result in correctly eliminating one variable (Other coefficients may be incorrect).
A1: Both
A1: Convincing justification but could be by finding the same \(\mathbf{r}\) from both equations
A1: Condone coordinates. Can be awarded even if previous A mark not awarded (i.e. if not checked third equation)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 5 \\ -2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 10 \\ 14 \\ 11 \end{pmatrix}\) | B1 | 3.1a |
| \(\mathbf{r} = \begin{pmatrix} 15 \\ -2 \\ 7 \end{pmatrix} + \nu\begin{pmatrix} 10 \\ 14 \\ 11 \end{pmatrix}\) | B1FT | 1.1 |
| \(\dfrac{x - 15}{10} = \dfrac{y + 2}{14} = \dfrac{z - 7}{11}\) | B1FT | 1.1 |
| [3] |
Notes
B1: Could be BC. SOI
B1FT: FT their point of intersection from (a) and their attempt at direction vector. SOI
Condone use of \(\lambda\) or \(\mu\).
No need to see “\(\mathbf{r} =\)”
Must be a recognisable attempt at a vector perpendicular to both \(L_1\) and \(L_2\)
B1FT: FT their vector equation. This is for correctly turning a vector equation into a cartesian one.
Correct equation here implies the other two marks.