AS June 2022 Paper 1 Q3
3.
| With respect to the right-hand rule, a rotation through \(\theta^\circ\) anticlockwise about the \(y\)-axis is represented by the matrix\[\begin{pmatrix}\cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta\end{pmatrix}\] |
The point \(P\) has coordinates (8, 3, 2)
The point \(Q\) is the image of \(P\) under the transformation reflection in the plane \(y = 0\)
The point \(R\) is the image of \(P\) under the transformation rotation through 120° anticlockwise about the \(y\)-axis, with respect to the right-hand rule.
| Scheme | Marks | AO |
|---|---|---|
| Coordinates of \(Q\) are \((8, -3, 2)\) | B1 | 2.2a |
| (1) |
Notes
B1: Coordinates of \(Q\) correctly stated, accept as a column vector.
| Scheme | Marks | AO |
|---|---|---|
| Coordinates of \(R\) are \(\begin{pmatrix}\cos 120^\circ & 0 & \sin 120^\circ\\ 0 & 1 & 0\\ -\sin 120^\circ & 0 & \cos 120^\circ\end{pmatrix}\begin{pmatrix}8\\ 3\\ 2\end{pmatrix} = \ldots\) or \(\begin{pmatrix}-0.5 & 0 & \dfrac{\sqrt{3}}{2}\\[8pt] 0 & 1 & 0\\[4pt] -\dfrac{\sqrt{3}}{2} & 0 & -0.5\end{pmatrix}\begin{pmatrix}8\\ 3\\ 2\end{pmatrix} = \ldots\) | M1 | 1.1a |
| So \(R\) is \(\left(-4 + \sqrt{3},\ 3,\ -4\sqrt{3} - 1\right)\) | A1 | 1.1b |
| (2) |
Notes
M1: Correct attempt to find coordinates of \(R\) using the given matrix with \(\theta = 120\). Must be multiplying in the correct way round. With no working two correct values or (−2.27, 3, −7.93) implies this mark.
A1: Correct exact coordinates as shown in scheme. Accept as a column vector. Cos 120 and sin 120 must have been evaluated.
| Scheme | Marks | AO |
|---|---|---|
| Finds the distance\[PR = \sqrt{\left(8 - \text{‘}\left(-4 + \sqrt{3}\right)\text{’}\right)^2 + (3 - \text{‘}3\text{’})^2 + \left(2 - \text{‘}\left(-4\sqrt{3} - 1\right)\text{’}\right)^2}\]Alternatively finds their \(\overrightarrow{PR}\) or their \(\overrightarrow{RP}\) then applies length of a vector formula.\[\sqrt{\left(12 - \sqrt{3}\right)^2 + \left(3 + 4\sqrt{3}\right)^2}\ \text{ or }\ \sqrt{\left(-12 + \sqrt{3}\right)^2 + \left(-3 - 4\sqrt{3}\right)^2}\] | M1 | 2.1 |
| \(= \sqrt{204}\ \left(= 2\sqrt{51}\right)\) cso | A1 | 1.1b |
| (2) |
Notes
M1: Applies the distance formula with the coordinates of \(P\) and their \(R\). Alternatively finds the vector \(\overrightarrow{PR}\) or \(\overrightarrow{RP}\) then applies length of a vector formula.
A1: Correct answer following correct coordinates of \(R\), must be a surd but need not be fully simplified.
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{PR}.\overrightarrow{PQ} = \left(-12 + \sqrt{3}, 0, -3 - 4\sqrt{3}\right).(0, -6, 0) = 0\) hence perpendicular | B1ft | 1.1b |
| (1) |
Notes
B1ft: Shows the dot product is zero between the vectors \(\overrightarrow{PR}\) and \(\overrightarrow{PQ}\) and draws the conclusion perpendicular. Accept with \(\pm\) vectors for each. Follow through as long as the vectors are of the correct form, so \(\overrightarrow{PR} = \begin{pmatrix}a\\ 0\\ b\end{pmatrix}\) and \(\overrightarrow{PQ} = \begin{pmatrix}0\\ c\\ 0\end{pmatrix}\)
Note They could state if vectors \(\overrightarrow{PR}\) and \(\overrightarrow{PQ}\) are perpendicular then \(\overrightarrow{PR}.\overrightarrow{PQ} = 0\) then shows \(\overrightarrow{PR}.\overrightarrow{PQ} = 0\) this is B1
| Scheme | Marks | AO |
|---|---|---|
| \(PQ\) is perpendicular to \(PR\) so Area \(= \dfrac{1}{2} \times PQ \times PR\) | M1 | 1.1b |
| \(= \dfrac{1}{2} \times 6 \times \sqrt{204} = 6\sqrt{51}\) cso | A1 | 1.1b |
| (2) | ||
| (8 marks) |
Notes
M1: Correct method for the area of the triangle, follow through on their coordinates of \(R\) and \(Q\). May see longer methods if they do not realise the triangle is right angled.
A1: For \(6\sqrt{51}\) cso following correct coordinates of \(R\)
Alternative 1
M1 Complete method to find the correct area
Finding all the lengths \(|PQ| = 6,\ |PR| = \sqrt{204} = 2\sqrt{51},\ |QR| = \sqrt{240} = 4\sqrt{15}\)
Use cosine rule to find an angle e.g. \(\cos PRQ = \dfrac{240 + 204 - 36}{2 \times \sqrt{240} \times \sqrt{204}} = \dfrac{\sqrt{85}}{10}\)
leading to \(PRQ = 22.7\ldots\) or \(\sin PRQ = \sqrt{1 - \left(\dfrac{\sqrt{85}}{10}\right)^2} = \ldots\left\{\dfrac{\sqrt{15}}{10}\right\}\)
Uses the area of the triangle \(= \dfrac{1}{2} \times \sqrt{240} \times \sqrt{204} \times \dfrac{\sqrt{15}}{10}\) or \(= \dfrac{1}{2} \times \sqrt{240} \times \sqrt{204} \times \sin 22.8\)
(Corrected from the printed mark scheme: the printed scheme has \(|PR| = \sqrt{240} = 4\sqrt{15}\) and \(|QR| = \sqrt{204} = 2\sqrt{51}\); these two lengths are the wrong way round, since part (c) gives \(|PR| = \sqrt{204}\).)
A1: For \(6\sqrt{51}\)
Alternative 2
M1: Uses \(\dfrac{1}{2}|\mathbf{a} \times \mathbf{b}|\) to find the required area
e.g. \(QP = \begin{pmatrix}0\\ 6\\ 0\end{pmatrix}\quad RP = \begin{pmatrix}12 - \sqrt{3}\\ 0\\ 3 + 4\sqrt{3}\end{pmatrix}\) cross product\[\begin{vmatrix}0 & 6 & 0\\ 12 - \sqrt{3} & 0 & 3 + 4\sqrt{3}\end{vmatrix} = 6\left(3 + 4\sqrt{3}\right)\mathbf{i} - 6\left(12 - \sqrt{3}\right)\mathbf{k}\]\[\text{Area} = \frac{1}{2}\sqrt{\left(6\left(3 + 4\sqrt{3}\right)\right)^2 + \left(-6\left(12 - \sqrt{3}\right)\right)^2} = \frac{1}{2}\sqrt{7344}\]
(Corrected from the printed mark scheme: the printed cross product is \(-6\left(12 - \sqrt{3}\right)\mathbf{i} + 6\left(3 + 4\sqrt{3}\right)\mathbf{k}\), with the \(\mathbf{i}\) and \(\mathbf{k}\) components the wrong way round. The area is unchanged.)
A1: For \(6\sqrt{51}\)