AS June 2025 Paper 1 Q6
6 The equations of two lines, \(l_1\) and \(l_2\), are \(l_1 : \mathbf{r} = \begin{pmatrix} 16 \\ -1 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -27 \\ -19 \end{pmatrix}\) and \(l_2 : \mathbf{r} = \begin{pmatrix} 3 \\ 10 \\ -10 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 10 \\ 10 \end{pmatrix}\).
\(O\) is the origin of the coordinate system. The point \(Q\) lies on the line segment \(OP\).
| Scheme | Marks | AO |
|---|---|---|
| \(16 + 2\lambda = 3 + \mu\) | M1 | 1.1 |
| \(16 + 2\lambda = 3 + \mu\) \(-1 - 27\lambda = 10 + 10\mu\) \(3 - 19\lambda = -10 + 10\mu\) | A1 | 1.1 |
| eg \(4 + 8\lambda = -20 \Rightarrow 8\lambda = -24 \Rightarrow \lambda = -3\) (subtracting 2nd from 3rd) | M1 | 1.1 |
| \(\Rightarrow 10\mu = 3 - 19(-3) + 10 = 70 \Rightarrow \mu = 7\) LHS \(= 16 + 2(-3) = 16 - 6 = 10\) RHS \(= 3 + 7 = 10\) | B1 | 1.1 |
| \(\therefore \mathbf{OP} = \begin{pmatrix} 16 \\ -1 \\ 3 \end{pmatrix} + (-3)\begin{pmatrix} 2 \\ -27 \\ -19 \end{pmatrix} = \begin{pmatrix} 10 \\ 80 \\ 60 \end{pmatrix}\) so \((10, 80, 60)\) | A1 | 1.1 |
| [5] |
Notes
M1: Equating \(x\), \(y\) or \(z\) components to derive an equation in \(\lambda\) and \(\mu\).
A1: Any two equations correct.
M1: Eliminating one unknown to solve for the other.
Could get eg \(430 = 101 + 47\mu\) leading to \(\mu = 7\).
Can be awarded for sight of \(\lambda = -3\) or \(\mu = 7\)
B1: Subbing back to find the other and checking for consistency in the unused equation.
Allow other valid ways of checking for consistency
Must see explicit check
A1: Condone just vector. \(\mathbf{OP} = \begin{pmatrix} 16 \\ -1 \\ 3 \end{pmatrix} + (-3)\begin{pmatrix} 2 \\ -27 \\ -19 \end{pmatrix} = \begin{pmatrix} 10 \\ 80 \\ 60 \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\therefore |\mathbf{OQ}|_{\max} = \sqrt{10^2 + 80^2 + 60^2} = \sqrt{10100}\) (since \(\mathbf{OQ} = \nu\begin{pmatrix} 10 \\ 80 \\ 60 \end{pmatrix}\), \(0 \leqslant \nu \leqslant 1\)) | M1 | 3.1a |
| \(= 100.498\ldots\) so \(OQ\) is probably \(\leqslant 100\) but there is a small chance that it is not. | A1FT | 2.2b |
| [2] |
Notes
A1FT: Allow 100.5 for \(OQ\)
Must follow an answer for 6(a) which gives \(OQ \gt 100\)
Allow “high chance that \(OQ\) is less than 100”
Allow “\(OQ\) can be greater than 100”
Accept \(\mathrm{P}(OQ \leqslant 100) = 0.995\)
Note: The probability that \(OQ \leqslant 100\) is \(0.995037\ldots\)
Do Not allow “\(OQ\) is greater than 100” or “claim is wrong”
Some indication that the claim might not be correct