| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Point on line satisfying distance or other condition |
| Difficulty | Standard +0.8 This is a multi-part 3D vectors question requiring distance calculation, angle between lines using dot product, and finding a point satisfying a distance condition. Part (c) requires solving a quadratic equation from |AC|=|AB|, which involves algebraic manipulation beyond routine exercises. The combination of techniques and the non-standard final part elevate this above typical textbook questions. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = (5-3)^2+(3--2)^2+(0-1)^2\) | M1 | Use \(\pm(\overrightarrow{OB}-\overrightarrow{OA})\) in sum of squares of components; allow one slip in difference |
| \(AB = \sqrt{30}\) | A1 | Accept 5.5 or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{bmatrix}2\\5\\-1\end{bmatrix}\cdot\begin{bmatrix}1\\0\\-3\end{bmatrix} = 2+3=5\) | M1 | \(\pm\overrightarrow{AB}\bullet\) direction \(l\) evaluated; condone one component error |
| \(5\) or \(-5\) | A1 | |
| \(\cos\theta = \frac{5}{\sqrt{30}\sqrt{10}}\) | B1F, M1 | F on either candidates' vectors; use \( |
| \(\theta = 73°\) | A1 | CAO (condone 73.2, 73.22 or 73.22...) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AC} = \begin{bmatrix}5+\lambda\\3\\-3\lambda\end{bmatrix}-\begin{bmatrix}3\\-2\\1\end{bmatrix}=\begin{bmatrix}2+\lambda\\5\\-1-3\lambda\end{bmatrix}\) | M1 | For \(\overrightarrow{OC}-\overrightarrow{OA}\) or \(\overrightarrow{OA}-\overrightarrow{OC}\) with \(\overrightarrow{OC}\) in terms of \(\lambda\); condone one component error |
| \((2+\lambda)^2+5^2+(-1-3\lambda)^2=30\) | A1 | |
| \(10\lambda^2+10\lambda=0\) | m1 | |
| \((\lambda=0\) or \()\,\lambda=-1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda=-1\Rightarrow C\) is \((4,3,3)\) | A1 | condone \(\begin{bmatrix}4\\3\\3\end{bmatrix}\) |
# Question 7:
## Part 7(a)
$AB^2 = (5-3)^2+(3--2)^2+(0-1)^2$ | M1 | Use $\pm(\overrightarrow{OB}-\overrightarrow{OA})$ in sum of squares of components; allow one slip in difference
$AB = \sqrt{30}$ | A1 | Accept 5.5 or better
## Part 7(b)
$\begin{bmatrix}2\\5\\-1\end{bmatrix}\cdot\begin{bmatrix}1\\0\\-3\end{bmatrix} = 2+3=5$ | M1 | $\pm\overrightarrow{AB}\bullet$ direction $l$ evaluated; condone one component error
$5$ or $-5$ | A1 |
$\cos\theta = \frac{5}{\sqrt{30}\sqrt{10}}$ | B1F, M1 | F on either candidates' vectors; use $|a||b|\cos\theta = a\bullet b$; values needed
$\theta = 73°$ | A1 | CAO (condone 73.2, 73.22 or 73.22...)
## Part 7(c)
$\overrightarrow{AC} = \begin{bmatrix}5+\lambda\\3\\-3\lambda\end{bmatrix}-\begin{bmatrix}3\\-2\\1\end{bmatrix}=\begin{bmatrix}2+\lambda\\5\\-1-3\lambda\end{bmatrix}$ | M1 | For $\overrightarrow{OC}-\overrightarrow{OA}$ or $\overrightarrow{OA}-\overrightarrow{OC}$ with $\overrightarrow{OC}$ in terms of $\lambda$; condone one component error
$(2+\lambda)^2+5^2+(-1-3\lambda)^2=30$ | A1 |
$10\lambda^2+10\lambda=0$ | m1 |
$(\lambda=0$ or $)\,\lambda=-1$ | A1 |
$(\lambda=0\Rightarrow(5,3,0)$ is $B)$
$\lambda=-1\Rightarrow C$ is $(4,3,3)$ | A1 | condone $\begin{bmatrix}4\\3\\3\end{bmatrix}$
---7 The coordinates of the points $A$ and $B$ are ( $3 , - 2,1$ ) and ( $5,3,0$ ) respectively.
The line $l$ has equation $\mathbf { r } = \left[ \begin{array} { l } 5 \\ 3 \\ 0 \end{array} \right] + \lambda \left[ \begin{array} { r } 1 \\ 0 \\ - 3 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance between $A$ and $B$.
\item Find the acute angle between the lines $A B$ and $l$. Give your answer to the nearest degree.
\item The points $B$ and $C$ lie on $l$ such that the distance $A C$ is equal to the distance $A B$. Find the coordinates of $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q7 [12]}}