| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from general external point to line |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question requiring routine techniques: finding a direction vector, writing a line equation, verifying a point lies on a line, and using perpendicularity conditions. Part (b)(ii) requires setting up a dot product equation and solving for a parameter, which is slightly above basic recall but still a textbook exercise with no novel insight required. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ -5 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}\) | M1 A1 | M1 for \(\pm(\overrightarrow{OA} - \overrightarrow{OB})\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((\mathbf{r}) = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}\) | B1 F | ft on \(\overrightarrow{AB}\); OE |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \\ 5 \end{pmatrix}\) | M1 | \(\mu\) found and verified or statement \(\mu = -3\) satisfies all components |
| \(1 + \mu = -2\), \(-1 - 2\mu = 5\), \(\mu = -3\) | ||
| \(\mu = -3\) alone B1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{PQ} = \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix} - \begin{pmatrix} -3 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 + 2\lambda \\ 8 - 4\lambda \\ -4 - 3\lambda \end{pmatrix}\) | M1 | \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}\) with \(\overrightarrow{OQ}\) in parametric form in terms of \(\lambda\) (can be inferred later) |
| or \(\begin{pmatrix} 6 + 2\lambda \\ 4 - 4\lambda \\ -7 - 3\lambda \end{pmatrix}\) | A1 | or \(\begin{pmatrix} 6 + 2\lambda \\ 4 - 4\lambda \\ -7 - 3\lambda \end{pmatrix}\) |
| \(\overrightarrow{PQ} \cdot \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}\) | M1 | \(\overrightarrow{PQ} \cdot \overrightarrow{OQ}\) with \(\overrightarrow{PQ}\) in terms of \(\lambda\) |
| \((4 + 2\lambda) + (-2)(-4 - 3\lambda) = 0\) | m1 | linear expression in \(\lambda\) equated to 0 |
| \(\lambda = -1.5\) | A1 F | ft on sign/arithmetic error in \(\overrightarrow{PQ}\) or equation |
| \(Q\) is \((-1, 11, 5.5)\) | A1 | CAO |
**9(a)(i)**
| $\overrightarrow{AB} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ -5 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}$ | M1 A1 | M1 for $\pm(\overrightarrow{OA} - \overrightarrow{OB})$ |
**9(a)(ii)**
| $(\mathbf{r}) = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}$ | B1 F | ft on $\overrightarrow{AB}$; OE |
**9(b)(i)**
| $\begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \\ 5 \end{pmatrix}$ | M1 | $\mu$ found and verified or statement $\mu = -3$ satisfies all components |
| $1 + \mu = -2$, $-1 - 2\mu = 5$, $\mu = -3$ | | |
| $\mu = -3$ alone B1 | A1 | |
**9(b)(ii)**
| $\overrightarrow{PQ} = \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix} - \begin{pmatrix} -3 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 + 2\lambda \\ 8 - 4\lambda \\ -4 - 3\lambda \end{pmatrix}$ | M1 | $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$ with $\overrightarrow{OQ}$ in parametric form in terms of $\lambda$ (can be inferred later) |
| or $\begin{pmatrix} 6 + 2\lambda \\ 4 - 4\lambda \\ -7 - 3\lambda \end{pmatrix}$ | A1 | or $\begin{pmatrix} 6 + 2\lambda \\ 4 - 4\lambda \\ -7 - 3\lambda \end{pmatrix}$ |
| $\overrightarrow{PQ} \cdot \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$ | M1 | $\overrightarrow{PQ} \cdot \overrightarrow{OQ}$ with $\overrightarrow{PQ}$ in terms of $\lambda$ |
| $(4 + 2\lambda) + (-2)(-4 - 3\lambda) = 0$ | m1 | linear expression in $\lambda$ equated to 0 |
| $\lambda = -1.5$ | A1 F | ft on sign/arithmetic error in $\overrightarrow{PQ}$ or equation |
| $Q$ is $(-1, 11, 5.5)$ | A1 | CAO |
**Total: 11 marks**
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**TOTAL: 75 marks**9 The points $A$ and $B$ lie on the line $l _ { 1 }$ and have coordinates $( 2,5,1 )$ and $( 4,1 , - 2 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the vector $\overrightarrow { A B }$.
\item Find a vector equation of the line $l _ { 1 }$, with parameter $\lambda$.
\end{enumerate}\item The line $l _ { 2 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } 1 \\ - 3 \\ - 1 \end{array} \right] + \mu \left[ \begin{array} { r } 1 \\ 0 \\ - 2 \end{array} \right]$.
\begin{enumerate}[label=(\roman*)]
\item Show that the point $P ( - 2 , - 3,5 )$ lies on $l _ { 2 }$.
\item The point $Q$ lies on $l _ { 1 }$ and is such that $P Q$ is perpendicular to $l _ { 2 }$. Find the coordinates of $Q$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q9 [11]}}