| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing basic 3D transformations and vector operations. Part (a) requires simple reflection (trivial), (b) applies a given rotation matrix with standard angle values, (c-d) use routine magnitude and dot product calculations, and (e) applies the standard triangle area formula. All parts follow standard procedures with no problem-solving insight required, making it slightly easier than average despite being multi-step. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors4.03f Linear transformations 3D: reflections and rotations about axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Coordinates of \(Q\) are \((8, -3, 2)\) | B1 | Accept as column vector |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(R = \begin{pmatrix}\cos120° & 0 & \sin120° \\ 0 & 1 & 0 \\ -\sin120° & 0 & \cos120°\end{pmatrix}\begin{pmatrix}8\\3\\2\end{pmatrix}\) | M1 | Correct matrix with \(\theta=120°\), multiplied correctly; 2 correct values or \((-2.27, 3, -7.93)\) implies mark |
| \(R\) is \(\left(-4+\sqrt{3},\ 3,\ -4\sqrt{3}-1\right)\) | A1 | Exact; \(\cos120°\) and \(\sin120°\) must be evaluated |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(PR = \sqrt{\!\left(8-(-4+\sqrt{3})\right)^2+(3-3)^2+\!\left(2-(-4\sqrt{3}-1)\right)^2}\) | M1 | Applies distance formula with \(P\) and their \(R\); or finds \(\overrightarrow{PR}\) then applies vector length |
| \(= \sqrt{(12-\sqrt{3})^2+(3+4\sqrt{3})^2} = \sqrt{204}\ \left(=2\sqrt{51}\right)\) | A1 | Correct answer following correct \(R\); must be a surd, need not be fully simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\overrightarrow{PR}\cdot\overrightarrow{PQ} = (-12+\sqrt{3},\ 0,\ -3-4\sqrt{3})\cdot(0,-6,0) = 0\), hence perpendicular | B1ft | Follow through their \(R\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(PQ \perp PR\), so Area \(= \dfrac{1}{2} \times PQ \times PR\) | M1 | |
| \(= \dfrac{1}{2} \times 6 \times \sqrt{204} = 6\sqrt{51}\) cso | A1 |
# Question 3:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Coordinates of $Q$ are $(8, -3, 2)$ | B1 | Accept as column vector |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $R = \begin{pmatrix}\cos120° & 0 & \sin120° \\ 0 & 1 & 0 \\ -\sin120° & 0 & \cos120°\end{pmatrix}\begin{pmatrix}8\\3\\2\end{pmatrix}$ | M1 | Correct matrix with $\theta=120°$, multiplied correctly; 2 correct values or $(-2.27, 3, -7.93)$ implies mark |
| $R$ is $\left(-4+\sqrt{3},\ 3,\ -4\sqrt{3}-1\right)$ | A1 | Exact; $\cos120°$ and $\sin120°$ must be evaluated |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $PR = \sqrt{\!\left(8-(-4+\sqrt{3})\right)^2+(3-3)^2+\!\left(2-(-4\sqrt{3}-1)\right)^2}$ | M1 | Applies distance formula with $P$ and their $R$; or finds $\overrightarrow{PR}$ then applies vector length |
| $= \sqrt{(12-\sqrt{3})^2+(3+4\sqrt{3})^2} = \sqrt{204}\ \left(=2\sqrt{51}\right)$ | A1 | Correct answer following correct $R$; must be a surd, need not be fully simplified |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\overrightarrow{PR}\cdot\overrightarrow{PQ} = (-12+\sqrt{3},\ 0,\ -3-4\sqrt{3})\cdot(0,-6,0) = 0$, hence perpendicular | B1ft | Follow through their $R$ |
## Part (e):
| Working | Mark | Guidance |
|---------|------|----------|
| $PQ \perp PR$, so Area $= \dfrac{1}{2} \times PQ \times PR$ | M1 | |
| $= \dfrac{1}{2} \times 6 \times \sqrt{204} = 6\sqrt{51}$ cso | A1 | |\begin{enumerate}
\item $\left[ \begin{array} { l } \text { With respect to the right-hand rule, a rotation through } \theta ^ { \circ } \text { anticlockwise about the } \\ y \text {-axis is represented by the matrix } \end{array} \right]$\\
$\left( \begin{array} { c c c } \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ - \sin \theta & 0 & \cos \theta \end{array} \right)$
\end{enumerate}
The point $P$ has coordinates (8, 3, 2)\\
The point $Q$ is the image of $P$ under the transformation reflection in the plane $y = 0$\\
(a) Write down the coordinates of $Q$
The point $R$ is the image of $P$ under the transformation rotation through $120 ^ { \circ }$ anticlockwise about the $y$-axis, with respect to the right-hand rule.\\
(b) Determine the exact coordinates of $R$\\
(c) Hence find $| \overrightarrow { P R } |$ giving your answer as a simplified surd.\\
(d) Show that $\overrightarrow { P R }$ and $\overrightarrow { P Q }$ are perpendicular.\\
(e) Hence determine the exact area of triangle $P Q R$, giving your answer as a surd in simplest form.
\hfill \mbox{\textit{Edexcel CP AS 2022 Q3 [8]}}