C4 June 2008 Q6
6. With respect to a fixed origin \(O\), the lines \(l_1\) and \(l_2\) are given by the equations\[\begin{aligned} l_1&:\quad \mathbf{r} = (-9\mathbf{i} + 10\mathbf{k}) + \lambda(2\mathbf{i} + \mathbf{j} - \mathbf{k})\\ l_2&:\quad \mathbf{r} = (3\mathbf{i} + \mathbf{j} + 17\mathbf{k}) + \mu(3\mathbf{i} - \mathbf{j} + 5\mathbf{k})\end{aligned}\]where \(\lambda\) and \(\mu\) are scalar parameters.
The point \(A\) has position vector \(5\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\).
The point \(B\) is the image of \(A\) after reflection in the line \(l_2\).
| Scheme | Marks |
|---|---|
| Lines meet where: \(\begin{pmatrix}-9\\0\\10\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}3\\1\\17\end{pmatrix} + \mu\begin{pmatrix}3\\-1\\5\end{pmatrix}\) | |
| Any two of \(\begin{aligned}\mathbf{i}:&\quad -9 + 2\lambda = 3 + 3\mu &&(1)\\ \mathbf{j}:&\quad \lambda = 1 - \mu &&(2)\\ \mathbf{k}:&\quad 10 - \lambda = 17 + 5\mu &&(3)\end{aligned}\) | M1 |
| (1) − 2(2) gives: \(-9 = 1 + 5\mu \Rightarrow \mu = -2\) | dM1 |
| (2) gives: \(\lambda = 1 - -2 = 3\) | A1 |
| \(\mathbf{r} = \begin{pmatrix}-9\\0\\10\end{pmatrix} + 3\begin{pmatrix}2\\1\\-1\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\1\\17\end{pmatrix} - 2\begin{pmatrix}3\\-1\\5\end{pmatrix}\) | ddM1 |
| Intersect at \(\mathbf{r} = \underline{\begin{pmatrix}-3\\3\\7\end{pmatrix}}\) or \(\mathbf{r} = \underline{-3\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}}\) | A1 |
| Either check \(\mathbf{k}\): \(\lambda = 3\): LHS \(= 10 - \lambda = 10 - 3 = 7\) \(\mu = -2\): RHS \(= 17 + 5\mu = 17 - 10 = 7\) (As LHS = RHS then the lines intersect.) | B1 |
| (6) |
Notes
M1: Need any two of these correct equations seen anywhere in part (a).
dM1: Attempts to solve simultaneous equations to find one of either \(\lambda\) or \(\mu\)
A1: Both \(\underline{\lambda = 3}\) & \(\underline{\mu = -2}\)
ddM1: Substitutes their value of either \(\lambda\) or \(\mu\) into the line \(l_1\) or \(l_2\) respectively. This mark can be implied by any two correct components of \((-3, 3, 7)\).
A1: \(\underline{\begin{pmatrix}-3\\3\\7\end{pmatrix}}\) or \(\underline{-3\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}}\) or \((-3, 3, 7)\)
B1: Either check that \(\lambda = 3,\ \mu = -2\) in a third equation or check that \(\lambda = 3,\ \mu = -2\) give the same coordinates on the other line. Conclusion not needed.
| Scheme | Marks |
|---|---|
| \(\mathbf{d}_1 = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\), \(\mathbf{d}_2 = 3\mathbf{i} - \mathbf{j} + 5\mathbf{k}\) | |
| As \(\mathbf{d}_1 \bullet \mathbf{d}_2 = \begin{pmatrix}2\\1\\-1\end{pmatrix} \bullet \begin{pmatrix}3\\-1\\5\end{pmatrix} = \underline{(2 \times 3) + (1 \times -1) + (-1 \times 5)} = 0\) | M1 |
| Then \(l_1\) is perpendicular to \(l_2\). | A1 |
| (2) |
Notes
M1: Dot product calculation between the two direction vectors: \(\underline{(2 \times 3) + (1 \times -1) + (-1 \times 5)}\) or \(\underline{6 - 1 - 5}\)
A1: Result ‘=0’ and appropriate conclusion
| Scheme | Marks |
|---|---|
| Equating \(\mathbf{i}\); \(-9 + 2\lambda = 5 \Rightarrow \lambda = 7\) | |
| \(\mathbf{r} = \begin{pmatrix}-9\\0\\10\end{pmatrix} + 7\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}5\\7\\3\end{pmatrix}\) (\(= \overrightarrow{OA}\). Hence the point A lies on \(l_1\).) | B1 |
| (1) |
Notes
B1: Substitutes candidate’s \(\lambda = 7\) into the line \(l_1\) and finds \(5\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\). The conclusion on this occasion is not needed.
| Scheme | Marks |
|---|---|
| Let \(\overrightarrow{OX} = -3\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}\) be point of intersection | |
| \(\overrightarrow{AX} = \overrightarrow{OX} - \overrightarrow{OA} = \underline{\begin{pmatrix}-3\\3\\7\end{pmatrix} - \begin{pmatrix}5\\7\\3\end{pmatrix}} = \begin{pmatrix}-8\\-4\\4\end{pmatrix}\) | M1ft \(\pm\) |
| \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OA} + 2\overrightarrow{AX}\) | |
| \(\overrightarrow{OB} = \begin{pmatrix}5\\7\\3\end{pmatrix} + 2\begin{pmatrix}-8\\-4\\4\end{pmatrix}\) | dM1ft |
| Hence, \(\overrightarrow{OB} = \underline{\begin{pmatrix}-11\\-1\\11\end{pmatrix}}\) or \(\overrightarrow{OB} = \underline{-11\mathbf{i} - \mathbf{j} + 11\mathbf{k}}\) | A1 |
| (3) | |
| (12 marks) |
Notes
M1ft \(\pm\): Finding the difference between their \(\overrightarrow{OX}\) (can be implied) and \(\overrightarrow{OA}\). \(\overrightarrow{AX} = \pm\left(\begin{pmatrix}-3\\3\\7\end{pmatrix} - \begin{pmatrix}5\\7\\3\end{pmatrix}\right)\)
dM1ft: \(\begin{pmatrix}5\\7\\3\end{pmatrix} + 2\left(\text{their } \overrightarrow{AX}\right)\)
A1: \(\underline{\begin{pmatrix}-11\\-1\\11\end{pmatrix}}\) or \(\underline{-11\mathbf{i} - \mathbf{j} + 11\mathbf{k}}\) or \(\underline{(-11, -1, 11)}\)