| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Triangle and parallelogram problems |
| Difficulty | Standard +0.3 This is a multi-part 3D vectors question requiring standard techniques: finding a vector between points, calculating angles using dot product, writing vector equations of lines, and using geometric constraints. While it has several parts and requires careful work with the trapezium condition, each step uses routine A-level methods without requiring novel insight or particularly challenging problem-solving. |
| 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 |
| \(\overrightarrow{AB} = (6-3, 0-(-2), 3-4) = \begin{pmatrix}3\\2\\-1\end{pmatrix}\) | B1 | |
| B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Direction of \(l_1\) is \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\) | M1 | Identify direction vector |
| \(\cos\theta = \frac{\overrightarrow{AB}\cdot\mathbf{d}}{ | \overrightarrow{AB} | |
| \(\overrightarrow{AB}\cdot\mathbf{d} = 6+(-2)+(-3)=1\) | A1 | |
| \( | \overrightarrow{AB} | =\sqrt{14}\), \( |
| \(\theta = \arccos\!\left(\frac{1}{14}\right) \approx 85.9°\) | A1 | Acute angle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(D = (3+4, -2-2, 4+6) = (7,-4,10)\) when \(\lambda=2\) | M1 | Find point D |
| \(\mathbf{r} = \begin{pmatrix}7\\-4\\10\end{pmatrix} + \mu\begin{pmatrix}3\\2\\-1\end{pmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | AD | \): find using \(\lambda=2\) scaled direction |
| \(AD = 2\sqrt{14}\), set \( | BC | = 2\sqrt{14}\) |
| \(C = (7+3\mu, -4+2\mu, 10-\mu)\) on \(l_2\) | M1 | |
| \( | BC | ^2 = (1+3\mu)^2+(−4+2\mu)^2+(7-\mu)^2 = 56\) |
| Solve to find \(\mu\) | A1 | |
| Coordinates of \(C\) | A1 |
## Question 8:
**Part (a)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = (6-3, 0-(-2), 3-4) = \begin{pmatrix}3\\2\\-1\end{pmatrix}$ | B1 | |
| | B1 | |
**Part (a)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Direction of $l_1$ is $\begin{pmatrix}2\\-1\\3\end{pmatrix}$ | M1 | Identify direction vector |
| $\cos\theta = \frac{\overrightarrow{AB}\cdot\mathbf{d}}{|\overrightarrow{AB}||\mathbf{d}|}$ | M1 | Correct formula |
| $\overrightarrow{AB}\cdot\mathbf{d} = 6+(-2)+(-3)=1$ | A1 | |
| $|\overrightarrow{AB}|=\sqrt{14}$, $|\mathbf{d}|=\sqrt{14}$ | A1 | |
| $\theta = \arccos\!\left(\frac{1}{14}\right) \approx 85.9°$ | A1 | Acute angle |
**Part (b)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D = (3+4, -2-2, 4+6) = (7,-4,10)$ when $\lambda=2$ | M1 | Find point D |
| $\mathbf{r} = \begin{pmatrix}7\\-4\\10\end{pmatrix} + \mu\begin{pmatrix}3\\2\\-1\end{pmatrix}$ | A1 | |
**Part (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|AD|$: find using $\lambda=2$ scaled direction | M1 | |
| $AD = 2\sqrt{14}$, set $|BC| = 2\sqrt{14}$ | A1 | |
| $C = (7+3\mu, -4+2\mu, 10-\mu)$ on $l_2$ | M1 | |
| $|BC|^2 = (1+3\mu)^2+(−4+2\mu)^2+(7-\mu)^2 = 56$ | M1 | |
| Solve to find $\mu$ | A1 | |
| Coordinates of $C$ | A1 | |8 The coordinates of the points $A$ and $B$ are $( 3 , - 2,4 )$ and $( 6,0,3 )$ respectively.\\
The line $l _ { 1 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } 3 \\ - 2 \\ 4 \end{array} \right] + \lambda \left[ \begin{array} { r } 2 \\ - 1 \\ 3 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the vector $\overrightarrow { A B }$.
\item Calculate the acute angle between $\overrightarrow { A B }$ and the line $l _ { 1 }$, giving your answer to the nearest $0.1 ^ { \circ }$.
\end{enumerate}\item The point $D$ lies on $l _ { 1 }$ where $\lambda = 2$. The line $l _ { 2 }$ passes through $D$ and is parallel to $A B$.
\begin{enumerate}[label=(\roman*)]
\item Find a vector equation of line $l _ { 2 }$ with parameter $\mu$.
\item The diagram shows a symmetrical trapezium $A B C D$, with angle $D A B$ equal to angle $A B C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{5fe2527a-33da-4076-b3fa-4cab545336ec-9_620_675_1197_726}
The point $C$ lies on line $l _ { 2 }$. The length of $A D$ is equal to the length of $B C$. Find the coordinates of $C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2011 Q8 [14]}}