FP3 June 2013 (R) Q4
4. The plane \(\Pi_1\) has vector equation \[\mathbf{r} = \begin{pmatrix}1 \\ -1 \\ 2\end{pmatrix} + s\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} + t\begin{pmatrix}1 \\ 2 \\ -2\end{pmatrix},\] where \(s\) and \(t\) are real parameters.
The plane \(\Pi_1\) is transformed to the plane \(\Pi_2\) by the transformation represented by the matrix \(\mathbf{T}\), where \[\mathbf{T} = \begin{pmatrix}2 & 0 & 3 \\ 0 & 2 & -1 \\ 0 & 1 & 2\end{pmatrix}\]
Find an equation of the plane \(\Pi_2\) in the form \(\mathbf{r} \cdot \mathbf{n} = p\) (9)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}2 & 0 & 3 \\ 0 & 2 & -1 \\ 0 & 1 & 2\end{pmatrix}\begin{pmatrix}1 + s + t \\ -1 + s + 2t \\ 2 - 2t\end{pmatrix}\) M1: Writes \(\Pi_1\) as a single vector A1: Correct statement | M1A1 |
| \(\begin{pmatrix}2 & 0 & 3 \\ 0 & 2 & -1 \\ 0 & 1 & 2\end{pmatrix}\begin{pmatrix}1 + s + t \\ -1 + s + 2t \\ 2 - 2t\end{pmatrix} = \begin{pmatrix}2 + 2s + 2t + 6 - 6t \\ -2 + 2s + 4t - 2 + 2t \\ -1 + s + 2t + 4 - 4t\end{pmatrix}\) M1: Correct attempt to multiply A1: Correct vector in any form | M1A1 |
| \(= \begin{pmatrix}8 + 2s - 4t \\ -4 + 2s + 6t \\ 3 + s - 2t\end{pmatrix}\) Correct simplified vector | B1 |
| \(\mathbf{r} = \begin{pmatrix}8 \\ -4 \\ 3\end{pmatrix} + s\begin{pmatrix}2 \\ 2 \\ 1\end{pmatrix} + t\begin{pmatrix}-4 \\ 6 \\ -2\end{pmatrix}\) | |
| \(\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ -4 & 6 & -2\end{vmatrix} = -10\mathbf{i} + 20\mathbf{k}\) M1: Attempts cross product of their direction vectors A1: Any multiple of \(-10\mathbf{i} + 20\mathbf{k}\) | M1A1 |
| \((8\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}) \cdot (\mathbf{i} - 2\mathbf{k}) = 8 - 6\) Attempt scalar product of their normal vector with their position vector | M1 |
| \(\mathbf{r} \cdot (\mathbf{i} - 2\mathbf{k}) = 2\) Correct equation (accept any correct equivalent e.g. \(\mathbf{r} \cdot (-10\mathbf{i} + 20\mathbf{k}) = -20\)) | A1 |
| (9) | |
| (9 marks) |