A2 June 2022 Q6
6. The points \(P\), \(Q\) and \(R\) have position vectors \(\begin{pmatrix}1\\ -2\\ 4\end{pmatrix}\), \(\begin{pmatrix}3\\ 1\\ -5\end{pmatrix}\) and \(\begin{pmatrix}2\\ 0\\ 3\end{pmatrix}\) respectively.
| Scheme | Marks | AO |
|---|---|---|
| Finds any two vectors \(\pm\overrightarrow{PQ},\ \pm\overrightarrow{PR}\) or \(\pm\overrightarrow{QR}\) \(\pm\begin{pmatrix}2\\ 3\\ -9\end{pmatrix}\) or \(\pm\begin{pmatrix}1\\ 2\\ -1\end{pmatrix}\) or \(\pm\begin{pmatrix}-1\\ -1\\ 8\end{pmatrix}\) | M1 | 1.1b |
| A correct equation for the plane \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\) \(\mathbf{a} = \begin{pmatrix}1\\ -2\\ 4\end{pmatrix}\) or \(\begin{pmatrix}3\\ 1\\ -5\end{pmatrix}\) or \(\begin{pmatrix}2\\ 0\\ 3\end{pmatrix}\) \(\mathbf{b}\) and \(\mathbf{c}\) are any two vectors from \(\pm\begin{pmatrix}2\\ 3\\ -9\end{pmatrix}\) or \(\pm\begin{pmatrix}1\\ 2\\ -1\end{pmatrix}\) or \(\pm\begin{pmatrix}-1\\ -1\\ 8\end{pmatrix}\) | A1 | 1.1b |
| (2) |
Notes
Accept alternative vector forms throughout.
M1: Finds any two vectors \(\pm\overrightarrow{PQ},\ \pm\overrightarrow{PR}\) or \(\pm\overrightarrow{QR}\) by subtracting relevant vectors. Two out of three values correct is sufficient to imply the correct method
A1: Any correct equation for the plane. Must start with \(\mathbf{r} = \ldots\)
| Scheme | Marks | AO |
|---|---|---|
| Forms two simultaneous equations by setting \(y = 0\) and \(z = 0\) e.g. \(-2 + 3\lambda + 2\mu = 0\) \(4 - 9\lambda - \mu = 0\) | M1 | 3.1a |
| Solves their simultaneous equations to find a value for \(\mu\) and a value for \(\lambda\) e.g. \(\left.\begin{aligned}-2 + 3\lambda + 2\mu &= 0\\ 4 - 9\lambda - \mu &= 0\end{aligned}\right\} \Rightarrow \lambda = 0.4,\ \mu = 0.4\) | dM1 | 1.1b |
| Uses their values of \(\mu\) and \(\lambda\) to find the \(x\) coordinate \(x = 1 + 2\lambda + \mu = 1 + 2(0.4) + (0.4) = \ldots\) | ddM1 | 1.1b |
| \((2.2,\ 0,\ 0)\) | A1 | 1.1b |
| (4) | ||
| (6 marks) |
Notes
M1: Uses their equation for the plane to form two simultaneous equations by setting \(y = 0\) and \(z = 0\)
dM1: Dependent on the previous method mark. Solves their simultaneous equations from the \(y\) and \(z\) coordinates to find a value for \(\mu\) and a value for \(\lambda\)
ddM1: Depends on both method marks. Uses their value for \(\mu\) and their value for \(\lambda\) to find the \(x\) coordinate
A1: Correct coordinates. Accept as a column vector or listed separately (\(y = 0\) and \(z = 0\) may be implied). Accept equivalent fractions, e.g. \(\left(\dfrac{11}{5}, 0, 0\right)\) or \(\left(\dfrac{33}{15}, 0, 0\right)\)
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix}2 & 3 & -9\\ 1 & 2 & -1\end{vmatrix} = (-3 + 18)\mathbf{i} - (-2 + 9)\mathbf{j} + (4 - 3)\mathbf{k}\) | M1 | 3.1a |
| \(\begin{pmatrix}15\\ -7\\ 1\end{pmatrix}.\begin{pmatrix}1\\ -2\\ 4\end{pmatrix} = 15 + 14 + 4 = 33\) leading to \(15x - 7y + z = 33\) | dM1 | 1.1b |
| \(15x - 7(0) + (0) = 33 \Rightarrow x = \ldots\) | ddM1 | 1.1b |
| \((2.2,\ 0,\ 0)\) | A1 | 1.1b |
| (4) |
M1: Finds the cross product of the vectors \(\mathbf{b}\) and \(\mathbf{c}\) for their plane. Allow one slip in expansion.
dM1: Finds the Cartesian equation of the plane
ddM1: Depends on both previous method marks. Sets \(y = 0\) and \(z = 0\) to find the \(x\) coordinate.
A1: Correct coordinates. Accept as a column vector or listed separately (\(y = 0\) and \(z = 0\) may be implied). Accept equivalent fractions, e.g. \(\left(\dfrac{11}{5}, 0, 0\right)\) or \(\left(\dfrac{33}{15}, 0, 0\right)\)