A2 June 2023 Paper 1 Q8
8 The points \(P\), \(Q\) and \(R\) have coordinates \((0, 2, 3)\), \((2, 0, 1)\) and \((1, 3, 0)\) respectively.
The acute angle between the line segments \(PQ\) and \(PR\) is \(\theta\).
The triangle \(PQR\) lies in the plane \(\Pi\).
The point \(S\) has coordinates \((5, 3, -1)\).
[The volume of a tetrahedron is \(\dfrac{1}{3} \times \text{area of base} \times \text{perpendicular height}\)] [4]
The tetrahedron \(PQRS\) is transformed to the tetrahedron \(P^{\prime}Q^{\prime}R^{\prime}S^{\prime}\) by a rotation about the \(y\)-axis.
The \(x\)-coordinate of \(S^{\prime}\) is \(2\sqrt{2}\).
| Scheme | Marks |
|---|---|
| \(\overrightarrow{PQ} = \begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix}, \overrightarrow{PR} = \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\) | B1 |
| \(\cos\theta = \dfrac{\begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}}{\sqrt{(2)^2 + (-2)^2 + (-2)^2} \times \sqrt{(1)^2 + (1)^2 + (-3)^2}}\) \(= \dfrac{6}{\sqrt{12}\sqrt{11}}\) | M1 |
| \(\Rightarrow \sin\theta = \left(\sqrt{1 - \left(\dfrac{6}{\sqrt{12}\sqrt{11}}\right)^2}\right) = \sqrt{\dfrac{8}{11}} = 2\sqrt{\dfrac{22}{11^2}} = \dfrac{2}{11}\sqrt{22}\) | A1 |
| [3] |
Notes
B1: Both correct or other way round
M1: Correct use of scalar product including correct method for magnitudes and dot product
A1: AG All working must be using exact forms
Alternative for M1 A1
| Scheme | Marks |
|---|---|
| \(\sin\theta = \dfrac{\left|\begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\right|}{\sqrt{(2)^2 + (-2)^2 + (-2)^2} \times \sqrt{(1)^2 + (1)^2 + (-3)^2}}\) | M1 |
| \(= \dfrac{4\left|\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right|}{\sqrt{12}\sqrt{11}} = \dfrac{4\sqrt{6}}{\sqrt{132}} = \dfrac{4}{\sqrt{22}} = \dfrac{2}{11}\sqrt{22}\) | A1 |
M1: Correct use of vector product to find \(|\sin\theta|\) including correct method for magnitudes, and correct method for \(\begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\) soi. Condone \(\sin\theta\) instead of \(|\sin\theta|\)
A1: AG.
Alternative method
| Scheme | Marks |
|---|---|
| \(|QR| = \sqrt{11}\), \(|RP| = \sqrt{11}\), \(|PQ| = \sqrt{12}\) | B1 |
| Cosine rule \(\Rightarrow \cos\theta = \dfrac{12 + 11 - 11}{2 \times \sqrt{11} \times \sqrt{12}} = \sqrt{\dfrac{3}{11}}\) | M1 |
| \(\Rightarrow \sin\theta = \sqrt{1 - \dfrac{3}{11}} = \sqrt{\dfrac{8}{11}} = \dfrac{2}{11}\sqrt{22}\) | A1 |
B1: Sides of triangle
M1: Cosine rule
Or
| Scheme | Marks |
|---|---|
| \(|QR| = \sqrt{11}\), \(|RP| = \sqrt{11}\), \(|PQ| = \sqrt{12}\) | B1 |
| Drop perpendicular from \(R\) to \(QP\) at \(M\) \(RM = \sqrt{11 - \left(\dfrac{1}{2}\sqrt{12}\right)^2} = \sqrt{8}\) | M1 |
| \(\Rightarrow \sin\theta = \sqrt{\dfrac{8}{11}} = \dfrac{2}{11}\sqrt{22}\) | A1 |
B1: Sides of triangle
M1: Recognises an isosceles triangle so uses a median line
| Scheme | Marks |
|---|---|
| \(\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix} = 4\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\) | B1 |
| \(\Rightarrow 2x + y + z = d\) | M1 |
| Sub for a point \(\Rightarrow d = 5 \quad \Rightarrow 2x + y + z = 5\) | A1 |
| [3] |
Notes
B1: BC or any other relevant vector product. Can be awarded if seen in (a)
M1: Attempt to use \(\mathbf{r}.\mathbf{n} = d\) to form linear equation.
ft their vector product.
A1: Substitutes coordinates of \(P\), \(Q\) or \(R\) to find \(d\). (multiples accepted)
| Scheme | Marks |
|---|---|
| \(D = \dfrac{\left|\begin{pmatrix} 5 \\ 3 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} - 5\right|}{\left|\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right|}\) | M1 |
| \(= \dfrac{7}{\sqrt{6}}\) | A1 |
| \(V = \dfrac{1}{3} \times \left(\dfrac{1}{2}\left|\overrightarrow{PQ}\right|\left|\overrightarrow{PR}\right|\sin\theta\right) \times D\) \(= \dfrac{1}{3} \times \dfrac{1}{2} \times 2\sqrt{3} \times \sqrt{11} \times \dfrac{2}{11}\sqrt{22} \times \dfrac{7}{\sqrt{6}}\) | M1 |
| \(= \dfrac{14}{3}\) | A1 |
| [4] |
Notes
M1: Uses the formula given in formula book, or any other complete method for the shortest distance, ft their \(\overrightarrow{PQ} \times \overrightarrow{PR}\).
A1: Correct shortest distance.
M1: Uses \(\frac{1}{2}\left|\overrightarrow{PQ}\right|\left|\overrightarrow{PR}\right|\sin\theta\) oe, multiplied by their D/3, or \(\frac{1}{2}\left|\begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\right|\) multiplied by their D/3, but must indicate that \(\frac{1}{2}\left|\begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\right|\) is the area of the base.
A1: AG.
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} \cos\phi & 0 & \sin\phi \\ 0 & 1 & 0 \\ -\sin\phi & 0 & \cos\phi \end{pmatrix}\begin{pmatrix} 5 \\ 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 2\sqrt{2} \\ \ldots \\ \ldots \end{pmatrix}\) \(\Rightarrow 5\cos\phi - \sin\phi = 2\sqrt{2}\) | M1 |
| \(\Rightarrow \sin^2\phi = \left(5\cos\phi - 2\sqrt{2}\right)^2\) \(\Rightarrow 1 - \cos^2\phi = 25\cos^2\phi - 20\sqrt{2}\cos\phi + 8\) \(\Rightarrow 26\cos^2\phi - 20\sqrt{2}\cos\phi + 7 = 0\) | M1 |
| \(\Rightarrow \cos\phi = \dfrac{\sqrt{2}}{2}, \ \dfrac{7\sqrt{2}}{26}\) | A1 |
| \(\sin\phi = 5\cos\phi - 2\sqrt{2} = \dfrac{\sqrt{2}}{2}, \ -\dfrac{17\sqrt{2}}{26}\) \(\begin{pmatrix} \cos\phi & 0 & \sin\phi \\ 0 & 1 & 0 \\ -\sin\phi & 0 & \cos\phi \end{pmatrix}\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos\phi \\ 3 \\ -\sin\phi \end{pmatrix}\) | M1 |
| \(\Rightarrow \begin{pmatrix} \frac{\sqrt{2}}{2} \\ 3 \\ -\frac{\sqrt{2}}{2} \end{pmatrix}, \ \begin{pmatrix} \frac{7\sqrt{2}}{26} \\ 3 \\ \frac{17\sqrt{2}}{26} \end{pmatrix}\) i.e. \(R^{\prime} = \left(\dfrac{\sqrt{2}}{2}, 3, -\dfrac{\sqrt{2}}{2}\right)\) or \(\left(\dfrac{7\sqrt{2}}{26}, 3, \dfrac{17\sqrt{2}}{26}\right)\) | A1 |
| [5] |
Notes
M1: For rotation matrix multiplied by \(\overrightarrow{OR}\) or \(\overrightarrow{OS}\).
M1: For a correct step to form quadratic equation in \(\sin\phi\) or \(\cos\phi\) only.
For reference: \(26\sin^2\phi + 4\sqrt{2}\sin\phi - 17 = 0\)
A1: Solves quadratic equation in \(\sin\phi\) or \(\cos\phi\). (Exact answers required)
M1: Uses their \(\sin\phi\) or \(\cos\phi\) with \(5\cos\phi - \sin\phi = 2\sqrt{2}\) to find \(\cos\phi\) or \(\sin\phi\) respectively, (even if only one root)
or from \(\sin\phi = \sqrt{1 - \cos^2\phi}\) or \(\cos\phi = \sqrt{1 - \sin^2\phi}\) (condone inclusion of \(\pm\)), or repeats previous method
and multiplies out the matrices.
A1: For both, and no others. Accept given as \(\overrightarrow{OR^{\prime}}\).
If M0M0A0M0A0, SCB1 for any \(R^{\prime}\) with \(y\)-coordinate \(= 3\), and no other \(y\)-coordinates.
Alternative method for first 3 marks
| Scheme | Marks |
|---|---|
| \(5\cos\phi - \sin\phi = 2\sqrt{2}\) | M1 |
| \(\Rightarrow 5\cos\phi - \sin\phi = \sqrt{26}\cos(\phi + \alpha) = 2\sqrt{2}\) where \(\tan\alpha = \dfrac{1}{5} \Rightarrow \sin\alpha = \dfrac{1}{\sqrt{26}}, \cos\alpha = \dfrac{5}{\sqrt{26}}\) | M1 |
| \(\sqrt{26}\sin(\phi + \alpha) = \pm\sqrt{26 - \left(2\sqrt{2}\right)^2} = \pm 3\sqrt{2}\) \(\cos\phi = \cos((\phi + \alpha) - \alpha) = \cos(\phi + \alpha)\cos\alpha + \sin(\phi + \alpha)\sin\alpha\) \(= \dfrac{2\sqrt{2}}{\sqrt{26}} \times \dfrac{5}{\sqrt{26}} \pm \dfrac{3\sqrt{2}}{\sqrt{26}} \times \dfrac{1}{\sqrt{26}}\) \(\Rightarrow \cos\phi = \dfrac{\sqrt{2}}{2}, \ \dfrac{7\sqrt{2}}{26}\) | A1 |
| [3] |
M1: For rotation matrix multiplied by \(\overrightarrow{OR}\) or \(\overrightarrow{OS}\).
M1: Expressing in the form \(R\cos(\phi + \alpha)\) or \(R\sin(\phi + \alpha)\) oe. Note that \(5\cos\phi - \sin\phi = \sqrt{26}\sin\left(\phi + \arctan\left(-\frac{1}{5}\right) + \pi\right)\)
A1: Solves for \(\sin\phi\) or \(\cos\phi\). For reference, \(\frac{\sqrt{2}}{2} \approx 0.707\), \(-\frac{17\sqrt{2}}{26} \approx -0.925\) and \(\frac{7\sqrt{2}}{26} \approx 0.381\).