WJEC Further Unit 1 2022 June — Question 8 7 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2022
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicLinear transformations
Type3D transformation matrices
DifficultyChallenging +1.2 This question requires knowledge of 3D rotation matrices (Further Maths content) and setting up an equation from the constraint that x' = y' after rotation. Students must construct the rotation matrix, apply it, equate coordinates, and manipulate surds to reach the required form. While it involves multiple steps and algebraic manipulation with surds, the rotation matrix is standard and the approach is methodical rather than requiring novel insight.
Spec4.03f Linear transformations 3D: reflections and rotations about axes
8. The point \(( x , y , z )\) is rotated through \(60 ^ { \circ }\) anticlockwise around the \(z\)-axis. After rotation, the value of the \(x\)-coordinate is equal to the value of the \(y\)-coordinate.
Show that \(y = ( a + \sqrt { b } ) x\), where \(a\), \(b\) are integers whose values are to be determined.
Question 8:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Rotation matrix: \(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}\)B1
\(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{1}{2}x - \frac{\sqrt{3}}{2}y \\ \frac{\sqrt{3}}{2}x + \frac{1}{2}y \\ z \end{pmatrix}\)M1, A1 Attempt to multiply; allow 1 error (possibly repeated)
\(x' = \frac{1}{2}x - \frac{\sqrt{3}}{2}y\), \(y' = \frac{\sqrt{3}}{2}x + \frac{1}{2}y\)
\(\frac{1}{2}x - \frac{\sqrt{3}}{2}y = \frac{\sqrt{3}}{2}x + \frac{1}{2}y\) leading to \(x(1-\sqrt{3}) = y(1+\sqrt{3})\)M1 FT their images matrix
\(y = \dfrac{x(1-\sqrt{3})}{1+\sqrt{3}}\)A1 cao
\(\dfrac{1-\sqrt{3}}{1+\sqrt{3}} = \dfrac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = \dfrac{4-2\sqrt{3}}{-2}\)M1 M0 no working; FT their \(y\) of equivalent difficulty e.g. \(y = \frac{x(a+\sqrt{b})}{c+\sqrt{d}}\)
\(y = (-2+\sqrt{3})x\)A1 
## Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Rotation matrix: $\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}$ | B1 | |
| $\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{1}{2}x - \frac{\sqrt{3}}{2}y \\ \frac{\sqrt{3}}{2}x + \frac{1}{2}y \\ z \end{pmatrix}$ | M1, A1 | Attempt to multiply; allow 1 error (possibly repeated) |
| $x' = \frac{1}{2}x - \frac{\sqrt{3}}{2}y$, $y' = \frac{\sqrt{3}}{2}x + \frac{1}{2}y$ | | |
| $\frac{1}{2}x - \frac{\sqrt{3}}{2}y = \frac{\sqrt{3}}{2}x + \frac{1}{2}y$ leading to $x(1-\sqrt{3}) = y(1+\sqrt{3})$ | M1 | FT their images matrix |
| $y = \dfrac{x(1-\sqrt{3})}{1+\sqrt{3}}$ | A1 | cao |
| $\dfrac{1-\sqrt{3}}{1+\sqrt{3}} = \dfrac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = \dfrac{4-2\sqrt{3}}{-2}$ | M1 | M0 no working; FT their $y$ of equivalent difficulty e.g. $y = \frac{x(a+\sqrt{b})}{c+\sqrt{d}}$ |
| $y = (-2+\sqrt{3})x$ | A1 | |

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8. The point $( x , y , z )$ is rotated through $60 ^ { \circ }$ anticlockwise around the $z$-axis. After rotation, the value of the $x$-coordinate is equal to the value of the $y$-coordinate.\\
Show that $y = ( a + \sqrt { b } ) x$, where $a$, $b$ are integers whose values are to be determined.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2022 Q8 [7]}}
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