| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Transformation mapping problems |
| Difficulty | Standard +0.3 This is a structured multi-part question on 3D transformations that guides students through standard procedures: identifying rotation angle from matrix entries (using basic trig values), finding matrix order, writing down a reflection matrix, applying composite transformations, and using dot product for angles. All parts are routine applications of Core Pure AS content with no novel problem-solving required. The calculations are straightforward with exact values (30°, 150° angles), making this slightly easier than average. |
| Spec | 4.03f Linear transformations 3D: reflections and rotations about axes4.03h Determinant 2x2: calculation4.03m det(AB) = det(A)*det(B)4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha = 210\) | B1 | Correct value; if more than one value stated, correct value must be clearly selected |
| Answer | Marks | Guidance |
|---|---|---|
| Require \(k \times\) their '210' divisible by 360 | M1 | Uses answer from (a) to find \(k\) such that \(k \times 210\) is divisible by 360; if answer to (a) is in radians, \(k \times \frac{7\pi}{6}\) divisible by \(2\pi\) |
| \(k = 12\) | A1 | Correct value must be using angle of 210; allow M1A1 for \(k=12\) following 210 or \(\frac{7\pi}{6}\); Note: angles 30, 150, 330 give \(k=12\) but is M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{N} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) | B1 | Correct matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{MA} = \begin{pmatrix} -\frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} -2 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} \sqrt{3}+2 \\ 1-2\sqrt{3} \\ 3 \end{pmatrix}\) then \(\mathbf{NMA} = \begin{pmatrix} \sqrt{3}+2 \\ 2\sqrt{3}-1 \\ 3 \end{pmatrix}\) | M1 | Complete method to find coordinates of \(B\); at least two correct terms for each multiplication; alternatively finds \(\mathbf{NM}\) (not \(\mathbf{MN}\)), look for 4 correct non-zero terms if no method shown, then multiplies to find coordinates of \(B\) |
| \(B\left(2+\sqrt{3},\ 2\sqrt{3}-1,\ 3\right)\) | A1* | Correct coordinates (condone vector); must have worked with exact values throughout; insufficient to just write \(\mathbf{NM}\begin{pmatrix}-2\\4\\3\end{pmatrix}\) without evidence of matrix multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = OA^2 + OB^2 - 2\,OA \cdot OB\cos AOB\) leading to \(AOB = 66.9°\), or \(\overrightarrow{OA}\cdot\overrightarrow{OB} = 1+6\sqrt{3} = \sqrt{29}\sqrt{29}\cos AOB\) | M1 | Identifies and applies appropriate strategy (cosine rule or scalar product); note \(AB^2 = 56-12\sqrt{3}\), \(AB = 3\sqrt{6}-\sqrt{2} = 5.93\ldots\) |
| \(AOB = 66.9°\) | A1* | Correct value from correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(AOB = \frac{1}{2}\sqrt{29}\sqrt{29}\sin 66.9°\) | M1 | Uses the given angle with \(\frac{1}{2}ab\sin C\) with appropriate \(a\), \(b\), \(C\); outside spec: uses cross product \(\frac{1}{2}\ |
| \(= 13.3\) | A1 | Correct area to 3 significant figures |
# Question 2:
## Part (a)
| $\alpha = 210$ | B1 | Correct value; if more than one value stated, correct value must be clearly selected |
## Part (b)
| Require $k \times$ their '210' divisible by 360 | M1 | Uses answer from (a) to find $k$ such that $k \times 210$ is divisible by 360; if answer to (a) is in radians, $k \times \frac{7\pi}{6}$ divisible by $2\pi$ |
| $k = 12$ | A1 | Correct value must be using angle of 210; allow M1A1 for $k=12$ following 210 or $\frac{7\pi}{6}$; **Note: angles 30, 150, 330 give $k=12$ but is M1A0** |
## Part (c)
| $\mathbf{N} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ | B1 | Correct matrix |
## Part (d)
| $\mathbf{MA} = \begin{pmatrix} -\frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} -2 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} \sqrt{3}+2 \\ 1-2\sqrt{3} \\ 3 \end{pmatrix}$ then $\mathbf{NMA} = \begin{pmatrix} \sqrt{3}+2 \\ 2\sqrt{3}-1 \\ 3 \end{pmatrix}$ | M1 | Complete method to find coordinates of $B$; at least two correct terms for each multiplication; alternatively finds $\mathbf{NM}$ (not $\mathbf{MN}$), look for 4 correct non-zero terms if no method shown, then multiplies to find coordinates of $B$ |
| $B\left(2+\sqrt{3},\ 2\sqrt{3}-1,\ 3\right)$ | A1* | Correct coordinates (condone vector); must have worked with exact values throughout; insufficient to just write $\mathbf{NM}\begin{pmatrix}-2\\4\\3\end{pmatrix}$ without evidence of matrix multiplication |
## Part (e)
| $AB^2 = OA^2 + OB^2 - 2\,OA \cdot OB\cos AOB$ leading to $AOB = 66.9°$, or $\overrightarrow{OA}\cdot\overrightarrow{OB} = 1+6\sqrt{3} = \sqrt{29}\sqrt{29}\cos AOB$ | M1 | Identifies and applies appropriate strategy (cosine rule or scalar product); note $AB^2 = 56-12\sqrt{3}$, $AB = 3\sqrt{6}-\sqrt{2} = 5.93\ldots$ |
| $AOB = 66.9°$ | A1* | Correct value from correct equation |
## Part (f)
| Area $AOB = \frac{1}{2}\sqrt{29}\sqrt{29}\sin 66.9°$ | M1 | Uses the given angle with $\frac{1}{2}ab\sin C$ with appropriate $a$, $b$, $C$; outside spec: uses cross product $\frac{1}{2}\|\overrightarrow{OA}\times\overrightarrow{OB}\|$ |
| $= 13.3$ | A1 | Correct area to 3 significant figures |
---\begin{enumerate}
\item $\left[ \begin{array} { l } \text { With respect to the right-hand rule, a rotation through } \theta ^ { \circ } \text { anticlockwise about } \\ \text { the } z \text {-axis is represented by the matrix } \\ \qquad \left( \begin{array} { c c c } \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right) \end{array} \right]$
\end{enumerate}
Given that the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { c c c }
- \frac { \sqrt { 3 } } { 2 } & \frac { 1 } { 2 } & 0 \\
- \frac { 1 } { 2 } & - \frac { \sqrt { 3 } } { 2 } & 0 \\
0 & 0 & 1
\end{array} \right)$$
represents a rotation through $\alpha ^ { \circ }$ anticlockwise about the $z$-axis with respect to the right-hand rule,\\
(a) determine the value of $\alpha$.\\
(b) Hence determine the smallest possible positive integer value of $k$ for which $\mathbf { M } ^ { k } = \mathbf { I }$
The $3 \times 3$ matrix $\mathbf { N }$ represents a reflection in the plane with equation $y = 0$\\
(c) Write down the matrix $\mathbf { N }$.
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 $\mathbf { N }$.\\
(d) Show that the coordinates of $B$ are $( 2 + \sqrt { 3 } , 2 \sqrt { 3 } - 1,3 )$
Given that $O$ is the origin,\\
(e) show that, to 3 significant figures, the size of angle $A O B$ is $66.9 ^ { \circ }$\\
(f) Hence determine the area of triangle $A O B$, giving your answer to 3 significant figures.
\hfill \mbox{\textit{Edexcel CP AS 2024 Q2 [10]}}