AS June 2024 Paper 1 Q2
2.
| With respect to the right-hand rule, a rotation through \(\theta^\circ\) anticlockwise about the \(z\)-axis is represented by the matrix\[\begin{pmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{pmatrix}\] |
Given that the matrix \(\mathbf{M}\), where
\[\mathbf{M} = \begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} & 0\\[4pt] -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} & 0\\[4pt] 0 & 0 & 1\end{pmatrix}\]represents a rotation through \(\alpha^\circ\) anticlockwise about the \(z\)-axis with respect to the right-hand rule,
The \(3 \times 3\) matrix \(\mathbf{N}\) represents a reflection in the plane with equation \(y = 0\)
The point \(A\) has coordinates \((-2, 4, 3)\)
The point \(B\) is the image of the point \(A\) under the transformation represented by matrix \(\mathbf{M}\) followed by the transformation represented by matrix \(\mathbf{N}\).
Given that \(O\) is the origin,
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha = 210\) | B1 | 1.1b |
| (1) |
Notes
B1: Correct value, check within the question. If more than one value is stated the correct value must clearly be selected.
| Scheme | Marks | AO |
|---|---|---|
| Require \(k \times\)their ‘210’ divisible by 360 | M1 | 1.1b |
| \(k = 12\) | A1 | 1.1b |
| (2) |
Notes
M1: Uses their answer from part (a) to determine a value for \(k\) so that \(k \times\)their 210 is divisible by 360. If their answer to part (a) is in radians \(k \times \text{their } \dfrac{7\pi}{6}\) is divisible by \(2\pi\).
A1: Correct value must be using an angle of 210. There may be no working but allow M1 A1 for \(k = 12\) following an answer of 210 or \(\dfrac{7\pi}{6}\) in part (a)
Note: an angle of 30, 150, 330 also gives \(\boldsymbol{k = 12}\) but is M1A0
| Scheme | Marks | AO |
|---|---|---|
| \(\{\mathbf{N} =\}\ \begin{pmatrix}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1\end{pmatrix}\) | B1 | 1.1b |
| (1) |
Notes
B1: Correct matrix
| Scheme | Marks | AO |
|---|---|---|
| \[\mathbf{MA} = \begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} & 0\\[4pt] -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} & 0\\[4pt] 0 & 0 & 1\end{pmatrix}\begin{pmatrix}-2\\ 4\\ 3\end{pmatrix} = \begin{pmatrix}\sqrt{3} + 2\\ 1 - 2\sqrt{3}\\ 3\end{pmatrix}\]\[\mathbf{NMA} = \begin{pmatrix}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}\sqrt{3} + 2\\ 1 - 2\sqrt{3}\\ 3\end{pmatrix} = \begin{pmatrix}\sqrt{3} + 2\\ 2\sqrt{3} - 1\\ 3\end{pmatrix} *\] Or \[\mathbf{NM} = \begin{pmatrix}1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} & 0\\[4pt] -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} & 0\\[4pt] 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} & 0\\[4pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} & 0\\[4pt] 0 & 0 & 1\end{pmatrix}\]\[\mathbf{NMA} = \begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} & 0\\[4pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} & 0\\[4pt] 0 & 0 & 1\end{pmatrix}\begin{pmatrix}-2\\ 4\\ 3\end{pmatrix} = \begin{pmatrix}\sqrt{3} + 2\\ 2\sqrt{3} - 1\\ 3\end{pmatrix} *\] | M1 | 1.1a |
| i.e. \(B\left(2 + \sqrt{3},\ 2\sqrt{3} - 1,\ 3\right)\) * | A1* | 1.1b |
| (2) |
Notes
M1: Complete method to find the coordinates of \(B\). Look for at least two correct terms for each multiplication, follow through when multiplying by N.
Alternatively finds NM (not MN), look for 4 correct non zero terms if no method is shown and then multiplies to find the coordinates of \(B\).
A1*: Correct coordinates (condone vector) Must have been working with exact values throughout. If working in decimals M1 A0
It is insufficient to just write \(\mathbf{NM}\begin{pmatrix}-2\\ 4\\ 3\end{pmatrix} = \begin{pmatrix}\sqrt{3} + 2\\ 2\sqrt{3} - 1\\ 3\end{pmatrix}\) there must be some evidence of matrix multiplication seen.
| Scheme | Marks | AO |
|---|---|---|
| \(AB^2 = OA^2 + OB^2 - 2OA\cdot OB\cos AOB\) \(\Rightarrow \left(4 + \sqrt{3}\right)^2 + \left(5 - 2\sqrt{3}\right)^2 = 29 + 29 - 2\sqrt{29}\cdot\sqrt{29}\cos AOB\) or \(OA\cdot OB = \begin{pmatrix}-2\\ 4\\ 3\end{pmatrix} \bullet \begin{pmatrix}2 + \sqrt{3}\\ 2\sqrt{3} - 1\\ 3\end{pmatrix} = 1 + 6\sqrt{3} = \sqrt{29}\sqrt{29}\cos AOB\) | M1 | 3.1a |
| \(AOB = 66.9^\circ\) * | A1* | 1.1b |
| (2) |
Notes
M1: Identifies and applies an appropriate strategy to find the required angle e.g. cosine rule or scalar product. Note: \(AB^2 = 56 - 12\sqrt{3}\) and \(AB = 3\sqrt{6} - \sqrt{2} = 5.93\ldots\)
A1*: Correct value from correct equation
| Scheme | Marks | AO |
|---|---|---|
| Area \(AOB = \dfrac{1}{2}\sqrt{29}\sqrt{29}\sin 66.9^\circ\) You may see this outside spec from candidates studying 8FM0 21\[\begin{aligned}&\frac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -2 & 4 & 3\\ 2 + \sqrt{3} & 2\sqrt{3} - 1 & 3\end{vmatrix}\\[8pt] &= \frac{1}{2}\left|\left[12 - 3\left(2\sqrt{3} - 1\right)\right]\mathbf{i} - \left[-6 - 3\left(2 + \sqrt{3}\right)\right]\mathbf{j} + \left[-2\left(2\sqrt{3} - 1\right) - 4\left(2 + \sqrt{3}\right)\right]\mathbf{k}\right|\\[8pt] &= \frac{1}{2}\left|\left(15 - 6\sqrt{3}\right)\mathbf{i} + \left(12 + 3\sqrt{3}\right)\mathbf{j} + \left(-6 - 8\sqrt{3}\right)\mathbf{k}\right|\\[8pt] &= \frac{1}{2}\sqrt{\left(15 - 6\sqrt{3}\right)^2 + \left(12 + 3\sqrt{3}\right)^2 + \left(-6 - 8\sqrt{3}\right)^2}\end{aligned}\] | M1 | 1.1b |
| = 13.3 cao | A1 | 1.1b |
| (2) | ||
| (10 marks) |
Notes
M1: Uses the given angle with \(\dfrac{1}{2}ab\sin C\) with appropriate \(a\), \(b\) and \(C\)
Outside spec: uses the cross product \(\dfrac{1}{2}\left|\overrightarrow{OA} \times \overrightarrow{OB}\right|\)
A1: Correct area to 3 significant figures