A2 June 2023 Paper 2 Q6
6 The equation of the plane \(\Pi\) is \(\mathbf{r} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\).
The point \(A\) has coordinates \((9, -7, 20)\).
The point \(F\) is the point of intersection between \(\Pi\) and the perpendicular from \(A\) to \(\Pi\).
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} \times \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -5 \\ -10 \\ 20 \end{pmatrix} = -5\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix}\) | B1 | 1.1 |
| \(\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}\ (= 2 - 12)\) | M1 | 1.1 |
| \(\cos\theta = \dfrac{-10}{\sqrt{1^2 + 2^2 + (-4)^2}\sqrt{2^2 + 3^2}}\) | M1 | 1.1 |
| \(\cos\theta = \dfrac{-10}{\sqrt{273}}\) so \(\theta = 127.2\) so required angle is \(52.8^\circ\) (1 dp) | A1 | 1.1 |
| [4] |
Notes
B1: Finding a normal to \(\Pi\). Any valid method; for example using \(\begin{pmatrix} 1 \\ a \\ b \end{pmatrix}\) and setting the dot product with both vectors in \(\Pi\) to 0.
M1: (1st) Any clear attempt to find the angle between the normals (can be implied by dotting the two normals together).
M1: (2nd) Using the definition of dot product to find \(\cos\theta\) in unsimplified numerical form. Could see modulus signs. This mark can be awarded after M0
A1: Or directly to answer. Final answer. awrt 0.921 rads
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} = -1 + 4 - 4\ (= -1)\) | M1 | 3.1a |
| \(\left(\begin{pmatrix} 9 \\ -7 \\ 20 \end{pmatrix} + \nu\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix}\right) \cdot \begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} = -1\) | M1 | 1.1 |
| \(9 - 14 - 80 + (1 + 4 + 16)\nu = -1 \Rightarrow \nu = 4\) | A1 | 1.1 |
| \(\mathbf{r}_F = \begin{pmatrix} 9 \\ -7 \\ 20 \end{pmatrix} + 4\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 13 \\ 1 \\ 4 \end{pmatrix}\) so \(F\) is \((13, 1, 4)\) | A1 | 1.1 |
| [4] |
Notes
M1: (1st) Dotting their normal and a point on \(\Pi\). Condone poor notation up to final A mark as long as method clear
M1: (2nd) Forming the equation of the line AF and intersecting with \(\Pi\) to find the value of the parameter for the PoI. Or M1 for using formula to find distance AF \(\left(= \frac{84}{\sqrt{21}}\right)\), and dividing this by magnitude their \(\mathbf{n}\).
A1: (2nd) Condone presentation as position vector of \(F\).
Alternative method
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 9 \\ -7 \\ 20 \end{pmatrix} + \nu\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\) so \(4\lambda - 2\mu - \nu = 10\) \(4\lambda + 3\mu - 2\nu = -9\) \(3\lambda + \mu + 4\nu = 19\) | M1 |
| \((\lambda = 2,\ \mu = -3)\ \nu = 4\) | M1 A1 |
| \(F\) is \((13, 1, 4)\) | A1 |
M1: (1st) Forms equations
M1 A1: BC. If not BC, then M1 for two equations in two unknowns
Alternative method 2
| Scheme | Marks |
|---|---|
| \(F\) is a point on \(\Pi\) so \(\overrightarrow{AF} = \overrightarrow{AO} + \overrightarrow{OF} = \begin{pmatrix} -9 \\ 7 \\ -20 \end{pmatrix} + \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\) \(= \begin{pmatrix} -10 \\ 9 \\ -19 \end{pmatrix} + \lambda\begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \mu\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\) \(\overrightarrow{AF} \cdot \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} = 0 \Rightarrow -61 + 41\lambda + 7\mu = 0\quad (41\lambda + 7\mu = 61)\) | M1 |
| \(\overrightarrow{AF} \cdot \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = 0 \Rightarrow 28 + 7\lambda + 14\mu = 0\quad (\lambda + 2\mu = -4)\) | M1 |
| \(\Rightarrow \lambda = 2,\ \mu = -3\) | A1 |
| \(\overrightarrow{OF} = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} - 3\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 13 \\ 1 \\ 4 \end{pmatrix}\) So \(F\) is \((13, 1, 4)\) | A1 |
M1: (1st) Using \(F\), a general point on \(\Pi\), equates \(\overrightarrow{AF} \cdot \mathbf{b}\) or \(\overrightarrow{AF} \cdot \mathbf{c}\) to 0
M1: (2nd) 2 equations in \(\lambda\) and \(\mu\)
A1: (1st) Solves (BC)
Alternative method 3
| Scheme | Marks |
|---|---|
| \(D\), perp dist from \(A\) to \(\Pi\) is given by \(D = \dfrac{\left|\begin{pmatrix} 9 \\ -7 \\ 20 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} - (-1)\right|}{\sqrt{1^2 + 2^2 + (-4)^2}} = \dfrac{|9 - 14 - 80 + 1|}{\sqrt{1 + 4 + 16}} = \dfrac{|-84|}{\sqrt{21}} = \dfrac{84}{21}\sqrt{21} = 4\sqrt{21}\) | M1 |
| \(\therefore \overrightarrow{AF} = 4\sqrt{21}\,\hat{\mathbf{n}} = 4\sqrt{21} \times \dfrac{1}{\sqrt{21}}\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} = 4\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ -16 \end{pmatrix}\) | A1 |
| \(\therefore \overrightarrow{OF} = \overrightarrow{OA} + \overrightarrow{AF} = \begin{pmatrix} 9 \\ -7 \\ 20 \end{pmatrix} + \begin{pmatrix} 4 \\ 8 \\ -16 \end{pmatrix} = \begin{pmatrix} 13 \\ 1 \\ 4 \end{pmatrix}\) So co-ords of \(F\) are \((13, 1, 4)\) (Or could see \(\overrightarrow{AF} = \lambda\begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix}\), \(\therefore \sqrt{\lambda^2 + (2\lambda)^2 + (-4\lambda)^2} = \sqrt{21}\lambda = 4\sqrt{21} \Rightarrow \lambda = 4\)) | M1 A1 |
M1: (1st) Finds perpendicular distance
A1: (1st) Finds normal vector from \(A\) to \(\Pi\)
M1 A1: Uses their normal vector to find \(F\)