| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Triangle and parallelogram areas |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: finding a position vector using parallelogram properties, calculating an angle using dot product, finding area using cross product, and applying geometric reasoning about trapezium area. All parts follow routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Attempts \(\overrightarrow{BA} = \mathbf{a} - \mathbf{b} = -2\mathbf{i} + 2\mathbf{j} - 8\mathbf{k}\) or \(\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = -4\mathbf{i} + 4\mathbf{j}\) | M1 | For attempting one of \(\mathbf{b}-\mathbf{a}\) or \(\mathbf{a}-\mathbf{b}\) or \(\mathbf{c}-\mathbf{b}\) or \(\mathbf{b}-\mathbf{c}\); must be correct for at least one component |
| \(\overrightarrow{OD} = \mathbf{a} - \mathbf{b} + \mathbf{c} = (-2\mathbf{i}+2\mathbf{j}-8\mathbf{k})+(-\mathbf{i}+3\mathbf{j}+6\mathbf{k}) = -3\mathbf{i}+5\mathbf{j}-2\mathbf{k}\) | M1 A1 | For attempting \(\mathbf{d} = \mathbf{a}-\mathbf{b}+\mathbf{c}\); must be correct for at least one component. cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overrightarrow{BA} = \mathbf{a}-\mathbf{b} = -2\mathbf{i}+2\mathbf{j}-8\mathbf{k}\) and \(\overrightarrow{BC} = \mathbf{c}-\mathbf{b} = -4\mathbf{i}+4\mathbf{j}\) | M1 | Uses correct pair of vectors \(\pm k\overrightarrow{BA}\) and \(\pm k\overrightarrow{BC}\); each must be correct for at least one component |
| \(\cos\theta = \dfrac{\begin{pmatrix}-2\\2\\-8\end{pmatrix}\cdot\begin{pmatrix}-4\\4\\0\end{pmatrix}}{\sqrt{(-2)^2+2^2+(-8)^2}\sqrt{(-4)^2+4^2}} = \dfrac{16}{\sqrt{72}\sqrt{32}} = \dfrac{1}{3}\) | dM1 A1 | Clear attempt to use dot product formula; dependent on previous M1. Allow arithmetical slips but process must be correct |
| Angle is \(1.23\) radians or \(70.5°\) | A1 | For \(\frac{1}{3}\) or \(-\frac{1}{3}\) or equivalent. cso for awrt \(70.5°\) or \(1.23\) rad |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Area \(= \sqrt{72}\sqrt{32}\sin\theta = 45.3\) or \(32\sqrt{2}\) | M1 A1 | Uses correct area formula for parallelogram. Obtains awrt \(45.3\); allow from angle of \(109.5°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Area \(= \dfrac{3}{2} \times 45.3 = 67.9\) or \(48\sqrt{2}\) | M1 A1 | Realises connection with part (c) and uses \(1.5\) times answer for area of \(ABCD\). awrt \(67.9\) |
## Question 14:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Attempts $\overrightarrow{BA} = \mathbf{a} - \mathbf{b} = -2\mathbf{i} + 2\mathbf{j} - 8\mathbf{k}$ or $\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = -4\mathbf{i} + 4\mathbf{j}$ | M1 | For attempting one of $\mathbf{b}-\mathbf{a}$ or $\mathbf{a}-\mathbf{b}$ or $\mathbf{c}-\mathbf{b}$ or $\mathbf{b}-\mathbf{c}$; must be correct for at least one component |
| $\overrightarrow{OD} = \mathbf{a} - \mathbf{b} + \mathbf{c} = (-2\mathbf{i}+2\mathbf{j}-8\mathbf{k})+(-\mathbf{i}+3\mathbf{j}+6\mathbf{k}) = -3\mathbf{i}+5\mathbf{j}-2\mathbf{k}$ | M1 A1 | For attempting $\mathbf{d} = \mathbf{a}-\mathbf{b}+\mathbf{c}$; must be correct for at least one component. cao |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overrightarrow{BA} = \mathbf{a}-\mathbf{b} = -2\mathbf{i}+2\mathbf{j}-8\mathbf{k}$ and $\overrightarrow{BC} = \mathbf{c}-\mathbf{b} = -4\mathbf{i}+4\mathbf{j}$ | M1 | Uses correct pair of vectors $\pm k\overrightarrow{BA}$ and $\pm k\overrightarrow{BC}$; each must be correct for at least one component |
| $\cos\theta = \dfrac{\begin{pmatrix}-2\\2\\-8\end{pmatrix}\cdot\begin{pmatrix}-4\\4\\0\end{pmatrix}}{\sqrt{(-2)^2+2^2+(-8)^2}\sqrt{(-4)^2+4^2}} = \dfrac{16}{\sqrt{72}\sqrt{32}} = \dfrac{1}{3}$ | dM1 A1 | Clear attempt to use dot product formula; dependent on previous M1. Allow arithmetical slips but process must be correct |
| Angle is $1.23$ radians or $70.5°$ | A1 | For $\frac{1}{3}$ or $-\frac{1}{3}$ or equivalent. cso for awrt $70.5°$ or $1.23$ rad |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Area $= \sqrt{72}\sqrt{32}\sin\theta = 45.3$ **or** $32\sqrt{2}$ | M1 A1 | Uses correct area formula for parallelogram. Obtains awrt $45.3$; allow from angle of $109.5°$ |
### Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Area $= \dfrac{3}{2} \times 45.3 = 67.9$ or $48\sqrt{2}$ | M1 A1 | Realises connection with part (c) and uses $1.5$ times answer for area of $ABCD$. awrt $67.9$ |
**Total: 11 marks**\begin{enumerate}
\item $A B C D$ is a parallelogram with $A B$ parallel to $D C$ and $A D$ parallel to $B C$. The position vectors of $A , B , C$, and $D$ relative to a fixed origin $O$ are $\mathbf { a } , \mathbf { b } , \mathbf { c }$ and $\mathbf { d }$ respectively.
\end{enumerate}
Given that
$$\mathbf { a } = \mathbf { i } + \mathbf { j } - 2 \mathbf { k } , \quad \mathbf { b } = 3 \mathbf { i } - \mathbf { j } + 6 \mathbf { k } , \quad \mathbf { c } = - \mathbf { i } + 3 \mathbf { j } + 6 \mathbf { k }$$
(a) find the position vector $\mathbf { d }$,\\
(b) find the angle between the sides $A B$ and $B C$ of the parallelogram,\\
(c) find the area of the parallelogram $A B C D$.
The point $E$ lies on the line through the points $C$ and $D$, so that $D$ is the midpoint of $C E$.\\
(d) Use your answer to part (c) to find the area of the trapezium $A B C E$.
\hfill \mbox{\textit{Edexcel C34 2017 Q14 [11]}}