C4 January 2008 Q6
6. The points \(A\) and \(B\) have position vectors \(2\mathbf{i} + 6\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} + 4\mathbf{j} + \mathbf{k}\) respectively.
The line \(l_1\) passes through the points \(A\) and \(B\).
A second line \(l_2\) passes through the origin and is parallel to the vector \(\mathbf{i} + \mathbf{k}\). The line \(l_1\) meets the line \(l_2\) at the point \(C\).
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OA} = \begin{pmatrix}2\\6\\-1\end{pmatrix}\) & \(\overrightarrow{OB} = \begin{pmatrix}3\\4\\1\end{pmatrix}\) | |
| \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}3\\4\\1\end{pmatrix} - \begin{pmatrix}2\\6\\-1\end{pmatrix} = \underline{\begin{pmatrix}1\\-2\\2\end{pmatrix}}\) | M1\(\pm\) A1 |
| (2) |
Notes
M1\(\pm\): Finding the difference between \(\overrightarrow{OB}\) and \(\overrightarrow{OA}\). A1: Correct answer.
| Scheme | Marks |
|---|---|
| \(l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\4\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}\) \(l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\-2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\4\\1\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\-2\end{pmatrix}\) | M1 A1ft aef |
| (2) |
Notes
M1: An expression of the form \((\text{vector}) \pm \lambda(\text{vector})\)
A1ft aef: \(\mathbf{r} = \overrightarrow{OA} \pm \lambda(\text{their } \overrightarrow{AB})\) or \(\mathbf{r} = \overrightarrow{OB} \pm \lambda(\text{their } \overrightarrow{AB})\) or \(\mathbf{r} = \overrightarrow{OA} \pm \lambda(\text{their } \overrightarrow{BA})\) or \(\mathbf{r} = \overrightarrow{OB} \pm \lambda(\text{their } \overrightarrow{BA})\) (\(\mathbf{r}\) is needed.)
| Scheme | Marks |
|---|---|
| \(l_2: \mathbf{r} = \begin{pmatrix}0\\0\\0\end{pmatrix} + \mu\begin{pmatrix}1\\0\\1\end{pmatrix} \Rightarrow \mathbf{r} = \mu\begin{pmatrix}1\\0\\1\end{pmatrix}\) | |
| \(\overrightarrow{AB} = \mathbf{d}_1 = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\), \(\mathbf{d}_2 = \mathbf{i} + 0\mathbf{j} + \mathbf{k}\) & \(\theta\) is angle | |
| \(\cos\theta = \dfrac{\overrightarrow{AB} \bullet \mathbf{d}_2}{\left(\left|\overrightarrow{AB}\right|.\left|\mathbf{d}_2\right|\right)} = \dfrac{\begin{pmatrix}1\\-2\\2\end{pmatrix} \bullet \begin{pmatrix}1\\0\\1\end{pmatrix}}{\left(\sqrt{(1)^2 + (-2)^2 + (2)^2}.\sqrt{(1)^2 + (0)^2 + (1)^2}\right)}\) | M1ft |
| \(\cos\theta = \dfrac{1 + 0 + 2}{\sqrt{(1)^2 + (-2)^2 + (2)^2}.\sqrt{(1)^2 + (0)^2 + (1)^2}}\) | A1ft |
| \(\cos\theta = \dfrac{3}{3.\sqrt{2}} \Rightarrow \underline{\theta = 45^\circ \text{ or } \tfrac{\pi}{4} \text{ or awrt } 0.79}.\) | A1 cao |
| (3) |
Notes
M1ft: Considers dot product between \(\mathbf{d}_2\) and their \(\overrightarrow{AB}\).
A1ft: Correct followed through expression or equation. This means that \(\cos\theta\) does not necessarily have to be the subject of the equation. It could be of the form \(3\sqrt{2}\cos\theta = 3\).
A1 cao: \(\underline{\theta = 45^\circ \text{ or } \tfrac{\pi}{4} \text{ or awrt } 0.79}\)
| Scheme | Marks |
|---|---|
| If \(l_1\) and \(l_2\) intersect then: \(\begin{pmatrix}2\\6\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix} = \mu\begin{pmatrix}1\\0\\1\end{pmatrix}\) | |
| \(\begin{aligned}\mathbf{i}:&\quad 2 + \lambda = \mu &&(1)\\ \mathbf{j}:&\quad 6 - 2\lambda = 0 &&(2)\\ \mathbf{k}:&\quad -1 + 2\lambda = \mu &&(3)\end{aligned}\) | M1ft |
| (2) yields \(\lambda = 3\) Any two yields \(\lambda = 3,\ \mu = 5\) | dM1 A1 |
| \(l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix} +3\begin{pmatrix}1\\-2\\2\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(\mathbf{r} = 5\begin{pmatrix}1\\0\\1\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) | A1 cso |
| (4) | |
| (11 marks) |
Notes
M1ft: Either seeing equation (2) written down correctly with or without any other equation or seeing equations (1) and (3) written down correctly.
dM1: Attempt to solve either equation (2) or simultaneously solve any two of the three equations to find … A1: either one of \(\lambda\) or \(\mu\) correct.
A1 cso: \(\underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(5\mathbf{i} + 5\mathbf{k}\). Fully correct solution & no incorrect values of \(\lambda\) or \(\mu\) seen earlier.
Aliter 6. (d) Way 2
| Scheme | Marks |
|---|---|
| If \(l_1\) and \(l_2\) intersect then: \(\begin{pmatrix}3\\4\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix} = \mu\begin{pmatrix}1\\0\\1\end{pmatrix}\) | |
| \(\begin{aligned}\mathbf{i}:&\quad 3 + \lambda = \mu &&(1)\\ \mathbf{j}:&\quad 4 - 2\lambda = 0 &&(2)\\ \mathbf{k}:&\quad 1 + 2\lambda = \mu &&(3)\end{aligned}\) | M1ft |
| (2) yields \(\lambda = 2\) Any two yields \(\lambda = 2,\ \mu = 5\) | dM1 A1 |
| \(l_1: \mathbf{r} = \begin{pmatrix}3\\4\\1\end{pmatrix} +2\begin{pmatrix}1\\-2\\2\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(\mathbf{r} = 5\begin{pmatrix}1\\0\\1\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) | A1 cso |
| (4) |
M1ft: Either seeing equation (2) written down correctly with or without any other equation or seeing equations (1) and (3) written down correctly.
dM1: Attempt to solve either equation (2) or simultaneously solve any two of the three equations to find … A1: either one of \(\lambda\) or \(\mu\) correct.
A1 cso: \(\underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(5\mathbf{i} + 5\mathbf{k}\). Fully correct solution & no incorrect values of \(\lambda\) or \(\mu\) seen earlier.
Aliter 6. (d) Way 3
| Scheme | Marks |
|---|---|
| If \(l_1\) and \(l_2\) intersect then: \(\begin{pmatrix}2\\6\\-1\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\-2\end{pmatrix} = \mu\begin{pmatrix}1\\0\\1\end{pmatrix}\) | |
| \(\begin{aligned}\mathbf{i}:&\quad 2 - \lambda = \mu &&(1)\\ \mathbf{j}:&\quad 6 + 2\lambda = 0 &&(2)\\ \mathbf{k}:&\quad -1 - 2\lambda = \mu &&(3)\end{aligned}\) | M1ft |
| (2) yields \(\lambda = -3\) Any two yields \(\lambda = -3,\ \mu = 5\) | dM1 A1 |
| \(l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix} -3\begin{pmatrix}-1\\2\\-2\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(\mathbf{r} = 5\begin{pmatrix}1\\0\\1\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) | A1 cso |
| (4) |
M1ft: Either seeing equation (2) written down correctly with or without any other equation or seeing equations (1) and (3) written down correctly.
dM1: Attempt to solve either equation (2) or simultaneously solve any two of the three equations to find … A1: either one of \(\lambda\) or \(\mu\) correct.
A1 cso: \(\underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(5\mathbf{i} + 5\mathbf{k}\). Fully correct solution & no incorrect values of \(\lambda\) or \(\mu\) seen earlier.
Aliter 6. (d) Way 4
| Scheme | Marks |
|---|---|
| If \(l_1\) and \(l_2\) intersect then: \(\begin{pmatrix}3\\4\\1\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\-2\end{pmatrix} = \mu\begin{pmatrix}1\\0\\1\end{pmatrix}\) | |
| \(\begin{aligned}\mathbf{i}:&\quad 3 - \lambda = \mu &&(1)\\ \mathbf{j}:&\quad 4 + 2\lambda = 0 &&(2)\\ \mathbf{k}:&\quad 1 - 2\lambda = \mu &&(3)\end{aligned}\) | M1ft |
| (2) yields \(\lambda = -2\) Any two yields \(\lambda = -2,\ \mu = 5\) | dM1 A1 |
| \(l_1: \mathbf{r} = \begin{pmatrix}3\\4\\1\end{pmatrix} -2\begin{pmatrix}-1\\2\\-2\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(\mathbf{r} = 5\begin{pmatrix}1\\0\\1\end{pmatrix} = \underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) | A1 cso |
| (4) |
M1ft: Either seeing equation (2) written down correctly with or without any other equation or seeing equations (1) and (3) written down correctly.
dM1: Attempt to solve either equation (2) or simultaneously solve any two of the three equations to find … A1: either one of \(\lambda\) or \(\mu\) correct.
A1 cso: \(\underline{\begin{pmatrix}5\\0\\5\end{pmatrix}}\) or \(5\mathbf{i} + 5\mathbf{k}\). Fully correct solution & no incorrect values of \(\lambda\) or \(\mu\) seen earlier.
Note: Be careful! \(\lambda\) and \(\mu\) are not defined in the question, so a candidate could interchange these or use different scalar parameters.