A2 October 2021 Paper 1 Q7
7. The plane \(\Pi\) has equation
\[\mathbf{r} = \begin{pmatrix}3\\ 3\\ 2\end{pmatrix} + \lambda\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix} + \mu\begin{pmatrix}2\\ 0\\ 1\end{pmatrix}\]where \(\lambda\) and \(\mu\) are scalar parameters.
The line \(l\) has equation
\[\mathbf{r} = \begin{pmatrix}4\\ -5\\ 2\end{pmatrix} + t\begin{pmatrix}1\\ 6\\ -3\end{pmatrix}\]where \(t\) is a scalar parameter.
The point \(A\) lies on \(l\).
Given that the shortest distance between \(A\) and \(\Pi\) is \(2\sqrt{29}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} = -2 + 6 - 4 = 0\) and \(\begin{pmatrix}2\\ 0\\ 1\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} = 4 + 0 - 4 = 0\) Alt: \(\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}\times\begin{pmatrix}2\\ 0\\ 1\end{pmatrix} = \begin{pmatrix}2 \times 1 - 1 \times 0\\ -(-1 \times 1 - 1 \times 2)\\ -1 \times 0 - 2 \times 2\end{pmatrix} = \ldots\) | M1 | 1.1b |
| As \(2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\) is perpendicular to both direction vectors (two non-parallel vectors) of \(\Pi\) then it must be perpendicular to \(\Pi\) | A1 | 2.2a |
| (2) |
Notes
M1: Attempts the scalar product of each direction vector and the vector \(2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\). Some numerical calculation is required, just “= 0” is insufficient. Alternatively, attempts the cross product (allow sign slips) with the two direction vectors.
A1: Shows that both scalar products = 0 (minimum \(-2 + 6 - 4 = 0\) and \(4 - 4 = 0\)) and makes a minimal conclusion with no erroneous statements. If using cross product, the calculation must be correct, and a minimal conclusion given with no erroneous statements.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} = \begin{pmatrix}3\\ 3\\ 2\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} \Rightarrow \ldots\) | M1 | 1.1a |
| \(2x + 3y - 4z = 7\) | A1 | 2.2a |
| (2) |
Notes
M1: Applies \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} = \begin{pmatrix}3\\ 3\\ 2\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} \Rightarrow \ldots\)
A1: \(2x + 3y - 4z = 7\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\left|2(4 + t) + 3(-5 + 6t) - 4(2 - 3t) - 7\right|}{\sqrt{2^2 + 3^2 + (-4)^2}} = 2\sqrt{29} \Rightarrow t = \ldots\) | M1 | 3.1a |
| \(t = -\dfrac{9}{8}\) and \(t = \dfrac{5}{2}\) | A1 | 1.1b |
| \(\boldsymbol{r} = \begin{pmatrix}4\\ -5\\ 2\end{pmatrix} - \dfrac{9}{8}\begin{pmatrix}1\\ 6\\ -3\end{pmatrix} = \ldots\) or \(\boldsymbol{r} = \begin{pmatrix}4\\ -5\\ 2\end{pmatrix} + \dfrac{5}{2}\begin{pmatrix}1\\ 6\\ -3\end{pmatrix} = \ldots\) | M1 | 1.1b |
| \(\left(\dfrac{23}{8}, -\dfrac{47}{4}, \dfrac{43}{8}\right)\) and \(\left(\dfrac{13}{2}, 10, -\dfrac{11}{2}\right)\) | A1 | 2.2a |
| (4) | ||
| (8 marks) |
Notes
M1: A fully correct method for finding a value of \(t\). Other methods are possible, but must be valid and lead to a value of \(t\). Examples of other methods:
- \(2\sqrt{29} = \pm\left(\dfrac{2(4 + t) + 3(-5 + 6t) - 4(2 - 3t)}{\sqrt{2^2 + 3^2 + (-4)^2}} - \dfrac{7}{\sqrt{29}}\right)\) using plane parallel to \(\Pi\) through origin and shortest distance from plane to origin.
- \(2(4 + t) + 3(-5 + 6t) - 4(2 - 3t) = 7 \Rightarrow t = t_i\) (\(t\) at intersection of line and plane) and
\(\sin\theta = \dfrac{(2, 3, -4)^T.(1, 6, -3)^T}{\sqrt{29}\sqrt{46}}\) (sine of angle between line and plane) followed by
\(\sin\theta = \dfrac{2\sqrt{29}}{k\sqrt{46}} \Rightarrow k = \ldots \Rightarrow t = t_i \pm k\)
A1: Correct values for \(t\). Both are required.
M1: Uses a value of \(t\) to find a set of coordinates for \(A\).
A1: Both correct sets of coordinates for \(A\).