A2 October 2021 Paper 2 Q3
3 The line \(l_1\) has equation \(\mathbf{r} = \begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\).
The plane \(\Pi\) has equation \(\mathbf{r}.\begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = 4\).
\(A\) is the point on \(l_1\) where \(\lambda = 1\).
\(l_2\) is the line with the following properties.
- \(l_2\) passes through \(A\)
- \(l_2\) is perpendicular to \(l_1\)
- \(l_2\) is parallel to \(\Pi\)
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\right).\begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = 4\) | M1 | 1.1 |
| \(2 + 15 - 9 + \lambda(6 - 10 + 6) = 4\) \(8 + 2\lambda = 4 \Rightarrow 2\lambda = -4 \Rightarrow \lambda = -2\) so | M1 | 1.1 |
| \(\mathbf{r} = \begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} + -2\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} -5 \\ -7 \\ 7 \end{pmatrix}\) | A1 | 1.1 |
| [3] |
Notes
M1: Substituting the expression for a point on the line into the equation of the plane
M1: Dotting out to form and solve equation in \(\lambda\)
A1: Condone coordinates
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}.\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}}{\sqrt{4 + 25 + 9}\sqrt{9 + 4 + 4}}\) soi \(= \dfrac{6 - 10 + 6}{\sqrt{38}\sqrt{17}} = \dfrac{2}{\sqrt{646}} = 0.07868\ldots\) | M1 | 1.1 |
| \(\theta =\) awrt \(85.5\ldots^\circ\) soi | A1 | 1.1 |
| \((\phi = 90^\circ - 85.48\ldots^\circ =)\) awrt \(4.51^\circ\) | A1 | 1.1 |
| [3] |
Notes
M1: BC. Using \(\cos\theta = \dfrac{\mathbf{a}.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)
May see \(\sin\phi = \dfrac{\mathbf{a}.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)
Or use of cross product
A1: (first) Can be implied by correct final answer
or \(1.49\ldots\) rads
A1: (second) or 0.0788 rads
| Scheme | Marks | AO |
|---|---|---|
| \(\lambda = 1 \Rightarrow \mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}\) | B1 | 3.1a |
| \(\mathbf{b} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = \begin{pmatrix} -16 \\ 5 \\ -19 \end{pmatrix}\) | M1 | 2.2a |
| So equation of \(l_2\) is \(\mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} -16 \\ 5 \\ -19 \end{pmatrix}\) oe | A1 | 1.1 |
| [3] |
Notes
M1: Method shown or at least two terms correctly evaluated
A1: Must be \(\mathbf{r} =\). Allow parameter \(\lambda\).