| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of parallelogram or trapezium using vectors |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: writing parallel line equations, reading coordinates from parametric form, finding angles using dot product, and calculating parallelogram area using cross product. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(l_2: \mathbf{r} = \begin{pmatrix}0\\0\\0\end{pmatrix}+\mu\begin{pmatrix}-1\\4\\3\end{pmatrix}\) | M1A1 | M1 for rhs with gradient \(k\times\begin{pmatrix}-1\\4\\3\end{pmatrix}\) containing point \(\begin{pmatrix}0\\0\\0\end{pmatrix}\); A1 correct with lhs \(\mathbf{r}=\); allow any scalar \(\mu\); condone another constant giving e.g. \(\mathbf{r}=\begin{pmatrix}1k\\-4k\\-3k\end{pmatrix}+\mu\begin{pmatrix}-1\\4\\3\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A=(2,3,-1)\) | B1 | Accept in vector notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B=(-1,15,8)\) | B1 | Accept in vector notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(\mathbf{a}=\begin{pmatrix}2\\3\\-1\end{pmatrix}\) and \(\mathbf{b}=k\begin{pmatrix}-1\\4\\3\end{pmatrix}\) or vice versa | M1 | Condone one sign slip only |
| \(\mathbf{a}\cdot\mathbf{b}=\ | \mathbf{a}\ | \ |
| \(\theta =\) awrt \(68.5°\) | A1 | Note \(1.2\) rad scores A0; allow awrt \(68.5°\) from \(180°-111.5°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB=\sqrt{3^2+12^2+9^2}\) or \(AB=3\times\sqrt{26}\) | M1 | Correct method for distance \(AB\); accept \(\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}\) or \(3\times\begin{vmatrix}-1\\4\\3\end{vmatrix}\) |
| Area \(= OA\times AB\times\sin(c) = \sqrt{14}\times 3\sqrt{26}\times\sin 68.5°\) | M1A1 | M1 for correct method for area of \(OABD\); accept two triangles method; awrt \(53\) (units²); accept exact answer \(9\sqrt{35}\) |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $l_2: \mathbf{r} = \begin{pmatrix}0\\0\\0\end{pmatrix}+\mu\begin{pmatrix}-1\\4\\3\end{pmatrix}$ | M1A1 | M1 for rhs with gradient $k\times\begin{pmatrix}-1\\4\\3\end{pmatrix}$ containing point $\begin{pmatrix}0\\0\\0\end{pmatrix}$; A1 correct with lhs $\mathbf{r}=$; allow any scalar $\mu$; condone another constant giving e.g. $\mathbf{r}=\begin{pmatrix}1k\\-4k\\-3k\end{pmatrix}+\mu\begin{pmatrix}-1\\4\\3\end{pmatrix}$ |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A=(2,3,-1)$ | B1 | Accept in vector notation |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B=(-1,15,8)$ | B1 | Accept in vector notation |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\mathbf{a}=\begin{pmatrix}2\\3\\-1\end{pmatrix}$ and $\mathbf{b}=k\begin{pmatrix}-1\\4\\3\end{pmatrix}$ or vice versa | M1 | Condone one sign slip only |
| $\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\|\mathbf{b}\|\cos\theta \Rightarrow -2+12-3=\sqrt{14}\sqrt{26}\cos\theta$ | dM1 | Correct method; condone one slip on $\mathbf{a}\cdot\mathbf{b}$; attempt at squaring and adding for $\|\mathbf{a}\|$ and $\|\mathbf{b}\|$ |
| $\theta =$ awrt $68.5°$ | A1 | Note $1.2$ rad scores A0; allow awrt $68.5°$ from $180°-111.5°$ |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB=\sqrt{3^2+12^2+9^2}$ or $AB=3\times\sqrt{26}$ | M1 | Correct method for distance $AB$; accept $\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$ or $3\times\begin{vmatrix}-1\\4\\3\end{vmatrix}$ |
| Area $= OA\times AB\times\sin(c) = \sqrt{14}\times 3\sqrt{26}\times\sin 68.5°$ | M1A1 | M1 for correct method for area of $OABD$; accept two triangles method; awrt $53$ (units²); accept exact answer $9\sqrt{35}$ |11. Relative to a fixed origin $O$, the line $l _ { 1 }$ is given by the equation
$$l _ { 1 } : \quad \mathbf { r } = \left( \begin{array} { r }
2 \\
3 \\
- 1
\end{array} \right) + \lambda \left( \begin{array} { r }
- 1 \\
4 \\
3
\end{array} \right)$$
where $\lambda$ is a scalar parameter.
The line $l _ { 2 }$ passes through the origin and is parallel to $l _ { 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for $l _ { 2 }$
The point $A$ and the point $B$ both lie on $l _ { 1 }$ with parameters $\lambda = 0$ and $\lambda = 3$ respectively.\\
Write down
\item \begin{enumerate}[label=(\roman*)]
\item the coordinates of $A$,
\item the coordinates of $B$.
\end{enumerate}\item Find the size of the acute angle between $O A$ and $l _ { 1 }$
Give your answer in degrees to one decimal place.
The point $D$ lies on $l _ { 2 }$ such that $O A B D$ is a parallelogram.
\item Find the area of $O A B D$, giving your answer to the nearest whole number.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q11 [10]}}