A2 October 2021 Paper 1 Q4
4 Points \(A\), \(B\) and \(C\) have coordinates \((4, 2, 0)\), \((1, 5, 3)\) and \((1, 4, -2)\) respectively.
The line \(l\) passes through \(A\) and \(B\).
\(M\) is the point on \(l\) that is closest to \(C\).
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix}\) oe | B1 | 1.1 |
| Equation of \(AB\) is \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix}\) oe | M1 | 1.1 |
| \(\Rightarrow 4 - x = y - 2 = z\) | A1 | 1.1 |
| [3] |
Notes
B1: soi
M1: their \(\begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix}\) soi \(\mathbf{r} =\) or \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\) is not required for M1
A1: Allow equivalent equations
e.g. \(\Rightarrow 1 - x = y - 5 = z - 3\) from using B
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix},\ \overrightarrow{OM} \text{ is } \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 - 3\lambda \\ 2 + 3\lambda \\ 3\lambda \end{pmatrix}\) | M1 | 3.1a |
| \(\overrightarrow{CM} = \begin{pmatrix} 4 - 3\lambda \\ 2 + 3\lambda \\ 3\lambda \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 - 3\lambda \\ 3\lambda - 2 \\ 3\lambda + 2 \end{pmatrix}\) | A1 | 1.1 |
| Perpendicular to \(\overrightarrow{AB} \Rightarrow \begin{pmatrix} 3 - 3\lambda \\ 3\lambda - 2 \\ 3\lambda + 2 \end{pmatrix}.\begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix} = 0\) \(\Rightarrow -9 + 9\lambda + 9\lambda - 6 + 9\lambda + 6 = 0\) | M1 | 1.1 |
| \(\Rightarrow 27\lambda = 9 \Rightarrow \lambda = \dfrac{1}{3}\). \(\Rightarrow\) Coordinates of \(M\) are \((3, 3, 1)\) | A1 | 1.1 |
| [4] |
Notes
M1: Attempt to find general point on AB to get vector CM. Can use \((1, 5, 3)\)
A1: Allow working throughout that uses e.g. \(\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\)
ft their vector from (i).
M1: Use of dot product to solve
A1: Do not accept a vector answer
Alternative method for last two marks
| Scheme | Marks |
|---|---|
| Minimise \(\left|\overrightarrow{CM}\right|^2 = \left|\begin{pmatrix} 3 - 3\lambda \\ 3\lambda - 2 \\ 3\lambda + 2 \end{pmatrix}\right|^2\) \(= (3 - 3\lambda)^2 + (3\lambda - 2)^2 + (3\lambda + 2)^2\) | M1 |
| \(\Rightarrow \lambda = \dfrac{1}{3} \Rightarrow\) Coordinates of \(M\) are \((3, 3, 1)\) | A1 |
| [4] |
M1: Express as a function of \(\lambda\) and minimise the quadratic in \(\lambda\)
| Scheme | Marks | AO |
|---|---|---|
| \(CM^2 = 2^2 + 1^2 + 3^2 = 14\) \(AB^2 = 3^2 + 3^2 + 3^2 = 27\) | B1 B1 | 1.1 1.1 |
| \(\Rightarrow \text{Area} = \dfrac{1}{2}\left|\overrightarrow{\mathrm{AB}}\right|.\left|\overrightarrow{\mathrm{CM}}\right|\) | M1 | 3.1a |
| \(= \dfrac{3}{2}\sqrt{42}\) | A1 | 1.1 |
| [4] |
Notes
B1: B1 for each distance ft their M
ft their AB
M1: Formula for area
ft their M
Alternative method 1
| Scheme | Marks |
|---|---|
| \(\text{Area} = \dfrac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{BC}\right| = \dfrac{1}{2}\left|\begin{pmatrix} 3 \\ -3 \\ -3 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 5 \end{pmatrix}\right| = \dfrac{1}{2}\left|\begin{pmatrix} -12 \\ -15 \\ 3 \end{pmatrix}\right|\) | M1 M1 A1 |
| \(= \dfrac{1}{2}\sqrt{12^2 + 15^2 + 3^2} = \dfrac{1}{2}\sqrt{378} = \dfrac{3}{2}\sqrt{42}\) | A1 |
M1: Formula for area
M1: Cross product
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\text{Area} = \dfrac{1}{2}\left|\overrightarrow{AB}\right|\left|\overrightarrow{BC}\right|\sin\theta\) where \(\overrightarrow{AB}.\overrightarrow{BC} = \left|\overrightarrow{AB}\right|\left|\overrightarrow{BC}\right|\cos\theta\) | M1 |
| \(\Rightarrow \cos\theta = \dfrac{-18}{\sqrt{27}\sqrt{26}} = \dfrac{-6}{\sqrt{78}}\) | A1 |
| \(\Rightarrow \sin\theta = \sqrt{1 - \dfrac{18}{39}} = \dfrac{\sqrt{21}}{\sqrt{39}}\) | M1 |
| \(\Rightarrow \text{Area} = \dfrac{1}{2}\sqrt{27}\sqrt{26}\dfrac{\sqrt{21}}{\sqrt{39}} = \dfrac{3}{2}\sqrt{42}\) | A1 |
| [4] |
M1: For use of dot product, formula for area
M1: Pythagoras to find \(\sin\theta\)
(Corrected from the printed mark scheme: Alternative method 2 is printed with \(\cos\theta = \dfrac{-12}{\sqrt{27}\sqrt{26}} = \dfrac{-4}{\sqrt{78}}\) and \(\sin\theta = \sqrt{1 - \dfrac{8}{39}} = \dfrac{\sqrt{31}}{\sqrt{39}}\), which do not give \(\frac{3}{2}\sqrt{42}\). With \(\overrightarrow{AB} = \begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 0 \\ -1 \\ -5 \end{pmatrix}\), \(\overrightarrow{AB}.\overrightarrow{BC} = -18\), as typed above.)