FP3 June 2014 (R) Q8
8. The plane \(\Pi_1\) has vector equation \(\mathbf{r} \cdot \begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix} = 5\)
The plane \(\Pi_2\) has vector equation \(\mathbf{r} \cdot \begin{pmatrix}-1 \\ 2 \\ 4\end{pmatrix} = 7\)
(a) Find a vector equation for the line of intersection of \(\Pi_1\) and \(\Pi_2\), giving your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\) where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors and \(\lambda\) is a scalar parameter. (6)
The plane \(\Pi_3\) has cartesian equation \[x - y + 2z = 31\]
(b) Using your answer to part (a), or otherwise, find the coordinates of the point of intersection of the planes \(\Pi_1\), \(\Pi_2\) and \(\Pi_3\) (3)
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ -1 & 2 & 4\end{vmatrix} = \begin{pmatrix}-2 \\ -11 \\ 5\end{pmatrix}\) M1: Attempt cross product of normal vectors. If method unclear, 2 components must be correct. A1: Correct vector | M1A1 |
| \(x = 0: \left(0, \tfrac{1}{2}, \tfrac{3}{2}\right),\ y = 0: \left(-\tfrac{1}{11}, 0, \tfrac{19}{11}\right),\ z = 0: \left(\tfrac{3}{5}, \tfrac{19}{5}, 0\right)\) M1: Attempt point on the line (\(x\), \(y\) and \(z\)). A1: Correct coordinates | M1A1 |
| \(\mathbf{r} = \tfrac{1}{2}\mathbf{j} + \tfrac{3}{2}\mathbf{k} + \lambda(2\mathbf{i} + 11\mathbf{j} - 5\mathbf{k})\) M1: Their point + \(\lambda\) their direction Dependent on both previous method marks. A: Correct equation (oe) | ddM1A1 |
| (6) |
Notes
Alternative 1
| Scheme | Marks |
|---|---|
| \(x = \dfrac{y - \frac{1}{2}}{\frac{11}{2}} = \dfrac{z - \frac{3}{2}}{-\frac{5}{2}}\) or \(\dfrac{x + \frac{1}{11}}{\frac{2}{11}} = y = \dfrac{z - \frac{19}{11}}{-\frac{5}{11}}\) or \(\dfrac{x - \frac{3}{5}}{-\frac{2}{5}} = \dfrac{y - \frac{19}{5}}{-\frac{11}{5}} = z\) M1: Correctly attempts cartesian equations of line A1: Correct equations | M1A1 |
| \(\left(0, \tfrac{1}{2}, \tfrac{3}{2}\right)\) or \(\left(-\tfrac{1}{11}, 0, \tfrac{19}{11}\right)\) or \(\left(\tfrac{3}{5}, \tfrac{19}{5}, 0\right)\) M1: Extracts position correctly A1: Correct position or \(\lambda(2\mathbf{i} + 11\mathbf{j} - 5\mathbf{k})\) M1: Extracts direction correctly A1: Correct direction | M1A1 |
| \(\mathbf{r} = \tfrac{1}{2}\mathbf{j} + \tfrac{3}{2}\mathbf{k} + \lambda(2\mathbf{i} + 11\mathbf{j} - 5\mathbf{k})\) M1: Their point + \(\lambda\) their direction Dependent on both previous method marks. A: Correct equation (oe) | ddM1A1 |
Alternative 2
| Scheme | Marks |
|---|---|
| \(x = \lambda \Rightarrow y = \dfrac{11\lambda + 1}{2},\ z = \dfrac{3 - 5\lambda}{2}\) (oe) M1: Obtains \(x\), \(y\) and \(z\) in terms of “\(\lambda\)” A1: Correct expressions | M1A1 |
| \(\left(0, \tfrac{1}{2}, \tfrac{3}{2}\right)\) or \(\lambda(2\mathbf{i} + 11\mathbf{j} - 5\mathbf{k})\) M1: Extracts position correctly A1: Correct position or M1: Extracts direction correctly A1: Correct direction | M1A1 |
| \(\mathbf{r} = \tfrac{1}{2}\mathbf{j} + \tfrac{3}{2}\mathbf{k} + \lambda(2\mathbf{i} + 11\mathbf{j} - 5\mathbf{k})\) M1: Their point + \(\lambda\) their direction Dependent on both previous method marks. A: Correct equation (oe) | ddM1A1 |
| Scheme | Marks |
|---|---|
| \(2\lambda - \left(\tfrac{1}{2} + 11\lambda\right) + 2\left(\tfrac{3}{2} - 5\lambda\right) = 31\) Substitutes into the third plane and solves for \(\lambda\) | M1 |
| \(\lambda = \tfrac{-3}{2}\) | |
| Planes intersect at \((-3, -16, 9)\) M1: Substitutes into their line A1: Correct coordinates | M1A1 |
| (3) | |
| (9 marks) |
Notes
Alternative
M1: Solves three simultaneous equations to obtain one value for \(x\) or \(y\) or \(x\)
M1: Solves three simultaneous equations to obtain values for \(x\), \(y\) and \(x\)
A1: Correct coordinates