FP3 June 2017 Q5
5. The plane \(\Pi_1\) has equation \(x - 2y - 3z = 5\) and the plane \(\Pi_2\) has equation \(6x + y - 4z = 7\)
The point \(P\) has coordinates \((2, 3, -1)\). The line \(l\) is perpendicular to \(\Pi_1\) and passes through the point \(P\). The line \(l\) intersects \(\Pi_2\) at the point \(Q\).
The plane \(\Pi_3\) passes through the point \(Q\) and is perpendicular to \(\Pi_1\) and \(\Pi_2\)
| Scheme | Marks |
|---|---|
| \(\Pi_1: x - 2y - 3z = 5,\quad \Pi_2: 6x + y - 4z = 7\) | |
| Way 1: \(\begin{pmatrix}1\\ -2\\ -3\end{pmatrix}.\begin{pmatrix}6\\ 1\\ -4\end{pmatrix} = 6 - 2 + 12\) | M1 |
| \(16 = \sqrt{1^2 + 2^2 + 3^2}\sqrt{6^2 + 1^2 + 4^2}\cos\theta\) \(\Rightarrow \cos\theta = \ldots\) | M1 |
| \(\cos\theta = \dfrac{16}{\sqrt{14}\sqrt{53}} \Rightarrow \theta = 54^\circ\) | A1 |
| (3) |
Notes
M1: Attempts scalar product of normal vectors allowing one slip. May be implied by a value of 16.
M1: Complete attempt to find \(\cos\theta\)
A1: Cao and do not isw. E.g. if they subsequently find 90 – 54 or 180 – 54, score A0. Do not allow 54.0.
(a) Way 2
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}1\\ -2\\ -3\end{pmatrix} \times \begin{pmatrix}6\\ 1\\ -4\end{pmatrix} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -2 & -3\\ 6 & 1 & -4\end{vmatrix} = \begin{pmatrix}11\\ -14\\ 13\end{pmatrix}\) | M1 |
| \(\sqrt{11^2 + 14^2 + 13^2} = \sqrt{1^2 + 2^2 + 3^2}\sqrt{6^2 + 1^2 + 4^2}\sin\theta\) \(\Rightarrow \sin\theta = \ldots\) | M1 |
| \(\sin\theta = \dfrac{9\sqrt{6}}{\sqrt{14}\sqrt{53}} \Rightarrow \theta = 54^\circ\) | A1 |
M1: Attempts cross product of normal vectors. 2 components should be correct if there is no working.
M1: Complete attempt to find \(\sin\theta\)
A1: Cao and do not isw. E.g. if they subsequently find 90 – 54 or 180 – 54, score A0. Do not allow 54.0.
| Scheme | Marks |
|---|---|
| \(\mathbf{PQ} = \begin{pmatrix}2\\ 3\\ -1\end{pmatrix} + \lambda\begin{pmatrix}1\\ -2\\ -3\end{pmatrix}\) or \(\begin{pmatrix}2 + \lambda\\ 3 - 2\lambda\\ -1 - 3\lambda\end{pmatrix}\) | M1 |
| \(6(2 + \lambda) + (3 - 2\lambda) - 4(-1 - 3\lambda) = 7\) \(\Rightarrow \lambda = \ldots\) | M1 |
| \(\lambda = -\dfrac{3}{4} \Rightarrow Q \textit{ is } \left(\dfrac{5}{4}, \dfrac{9}{2}, \dfrac{5}{4}\right)\) | M1A1 |
| (4) |
Notes
M1: Attempt parametric form of \(\mathbf{PQ}\) by using the point \(P\) and the normal to \(\Pi_1\)
M1: Substitutes parametric form of \(\mathbf{PQ}\) into the equation of \(\Pi_2\) and solves for \(\lambda\)
M1: Uses their value of \(\lambda\) in their \(\mathbf{PQ}\) equation
A1: Correct coordinates or vector.
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}1\\ -2\\ -3\end{pmatrix} \times \begin{pmatrix}6\\ 1\\ -4\end{pmatrix} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -2 & -3\\ 6 & 1 & -4\end{vmatrix} = \begin{pmatrix}11\\ -14\\ 13\end{pmatrix}\) | M1A1 |
| \(\begin{pmatrix}11\\ -14\\ 13\end{pmatrix}.\begin{pmatrix}\frac{5}{4}\\ \frac{9}{2}\\ \frac{5}{4}\end{pmatrix} = \ldots\) or \(\begin{pmatrix}11\\ -14\\ 13\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -1\end{pmatrix} = \ldots\) | M1 |
| \(\mathbf{r}.\begin{pmatrix}11\\ -14\\ 13\end{pmatrix} = -33\) | A1 |
| (4) | |
| (11 marks) |
Notes
M1: Attempt cross product between normals
A1: Correct normal vector (any multiple)
Alternative
\(x - 2y - 3z = 0,\ 6x + y - 4z = 0: x = 1 \Rightarrow y = -\dfrac{14}{11}, z = \dfrac{13}{11}\)
\(\Rightarrow \mathbf{n} = \begin{pmatrix}11\\ -14\\ 13\end{pmatrix}\)
M1: Solves \(x - 2y - 3z = 0,\ 6x + y - 4z = 0\) to obtain values for \(x\), \(y\) and \(z\)
A1: Correct vector (or values)
M1: Attempt scalar product between their normal and their \(\mathbf{OQ}\) or \(\mathbf{OP}\). Must obtain a value.
A1: Any multiple e.g. \(\mathbf{r}.\begin{pmatrix}11k\\ -14k\\ 13k\end{pmatrix} = -33k\quad (k \neq 0)\)
Note that if they use the intersection with \(\Pi_1\) \(\left(\dfrac{17}{7}, \dfrac{15}{7}, \dfrac{-16}{7}\right)\) for \(Q\) allow all the marks to score in (c).