| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring routine techniques: finding a position vector by substitution, showing lines intersect by equating components and solving simultaneously, and using parallelogram properties with vector addition. All parts follow predictable methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OB} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) | B1 | PI |
| \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \\ 0 \end{pmatrix}\) | M1 A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(4 + 2\lambda = -1 + \mu\); \(-3 = 3 - 2\mu\); \(2 + \lambda = 4 - \mu\) | M1 | Set up 3 equations |
| \(-6 = -2\mu\), \(\mu = 3\); \(\lambda = 4 - 3 - 2 = -1\); \(\lambda = -1 - 2 = -2\) | m1 | |
| \(4 + 2\lambda = 4 - 2 = 2\) | A1 | |
| \(-1 + \mu = -1 + 3 = 2\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\) is \((2, -3, 1)\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC}\) | M1 | Or \(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}\) or \(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{PA}\) |
| \(= \overrightarrow{OA} + \overrightarrow{PB}\) | M1 | \(\overrightarrow{OA} + \overrightarrow{PB}\) in components |
| \(\overrightarrow{OC} = \begin{pmatrix} 4 \\ -3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1-2 \\ -1-(-3) \\ 2-1 \end{pmatrix}\) | M1 | \(\overrightarrow{OA} + \overrightarrow{PB}\) in components |
| \(C\) is \((3, -1, 3)\) | A1 | |
| Or \(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}\) or \(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{AP}\) | M1 | \(\overrightarrow{OB} + \overrightarrow{AP}\) in components |
| \(\overrightarrow{OC} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \begin{pmatrix} 2-4 \\ -3-(-3) \\ 1-2 \end{pmatrix}\) | M1 | \(\overrightarrow{OB} + \overrightarrow{AP}\) in components |
| \(C\) is \((-1, -1, 1)\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AP} = \overrightarrow{BC}\) | M1 | |
| \( | \overrightarrow{AP} | = |
| \(\overrightarrow{BC} = k\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}\), \( | \overrightarrow{BC} | = \sqrt{k}\sqrt{5}\), so \(k = \pm 1\) |
| \(\overrightarrow{OC} = \overrightarrow{OB} + k\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}\) or \(\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}\) | M1 | Either |
| \(= \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}\) or \(\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}\) | A1 | 4 marks |
**7(a)**
$\overrightarrow{OB} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ | B1 | PI
$\overrightarrow{AB} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \\ 0 \end{pmatrix}$ | M1 A1 | 3 marks | Use $\pm(\overrightarrow{OB} - \overrightarrow{OA})$
**7(b)(i)**
$4 + 2\lambda = -1 + \mu$; $-3 = 3 - 2\mu$; $2 + \lambda = 4 - \mu$ | M1 | Set up 3 equations
$-6 = -2\mu$, $\mu = 3$; $\lambda = 4 - 3 - 2 = -1$; $\lambda = -1 - 2 = -2$ | m1
$4 + 2\lambda = 4 - 2 = 2$ | A1
$-1 + \mu = -1 + 3 = 2$ | A1 | 4 marks | Independent check with conclusion: minimum "intersect"
**7(b)(ii)**
$P$ is $(2, -3, 1)$ | B1 | 1 mark
**7(c)**
$\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC}$ | M1 | Or $\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}$ or $\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{PA}$
$= \overrightarrow{OA} + \overrightarrow{PB}$ | M1 | $\overrightarrow{OA} + \overrightarrow{PB}$ in components
$\overrightarrow{OC} = \begin{pmatrix} 4 \\ -3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1-2 \\ -1-(-3) \\ 2-1 \end{pmatrix}$ | M1 | $\overrightarrow{OA} + \overrightarrow{PB}$ in components
$C$ is $(3, -1, 3)$ | A1
Or $\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}$ or $\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{AP}$ | M1 | $\overrightarrow{OB} + \overrightarrow{AP}$ in components
$\overrightarrow{OC} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \begin{pmatrix} 2-4 \\ -3-(-3) \\ 1-2 \end{pmatrix}$ | M1 | $\overrightarrow{OB} + \overrightarrow{AP}$ in components
$C$ is $(-1, -1, 1)$ | A1 | 4 marks
**7(c) Alternative**
$\overrightarrow{AP} = \overrightarrow{BC}$ | M1
$|\overrightarrow{AP}| = |\overrightarrow{BC}| = \sqrt{(2-4)^2 + (-3-(-3))^2 + (1-2)^2} = \sqrt{5}$ | M1
$\overrightarrow{BC} = k\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$, $|\overrightarrow{BC}| = \sqrt{k}\sqrt{5}$, so $k = \pm 1$ | A1* | For $k=1$ and $k=-1$
$\overrightarrow{OC} = \overrightarrow{OB} + k\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ or $\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ | M1 | Either
$= \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$ or $\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}$ | A1 | 4 marks | Both
---The point $A$ has coordinates $(4, -3, 2)$.
The line $l_1$ passes through $A$ and has equation $\mathbf{r} = \begin{bmatrix} 4 \\ -3 \\ 2 \end{bmatrix} + \lambda \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$.
The line $l_2$ has equation $\mathbf{r} = \begin{bmatrix} -1 \\ 3 \\ 4 \end{bmatrix} + \mu \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}$.
The point $B$ lies on $l_2$ where $\mu = 2$.
\begin{enumerate}[label=(\alph*)]
\item Find the vector $\overrightarrow{AB}$. [3 marks]
\item
\begin{enumerate}[label=(\roman*)]
\item Show that the lines $l_1$ and $l_2$ intersect. [4 marks]
\item The lines $l_1$ and $l_2$ intersect at the point $P$. Find the coordinates of $P$. [1 mark]
\end{enumerate}
\item The point $C$ lies on a line which is parallel to $l_1$ and which passes through the point $B$. The points $A$, $B$, $C$ and $P$ are the vertices of a parallelogram.
Find the coordinates of the two possible positions of the point $C$. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2010 Q7 [12]}}