| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| 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 structured multi-part question with clear signposting ('show that') that guides students through standard vector techniques: distance formula, angle between lines using dot product, and finding perpendicular from point to line. Each part builds on the previous one with routine calculations. While it requires multiple steps, the methods are textbook-standard for C4 with no novel insight needed, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d 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 |
|---|---|---|
| (a) \( | AB | = \sqrt{(2-1)^2 + (-1-4)^2 + (3-2)^2} = \sqrt{27} = 3\sqrt{3}\) |
| (b) \(\begin{bmatrix} -5 \\ \square \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = 1 + 5 + 1\) | M1 | Their \(\overrightarrow{AB} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\) |
| \(7 = 3\sqrt{3}\sqrt{3}\cos\theta\), \(\cos\theta = \frac{7}{9}\) | M1, A1 | Their scalar product \(= |
| (c)(i) \(\overrightarrow{OP} = \begin{bmatrix} 2 + p \\ -1 - p \\ 3 + p \end{bmatrix}\), \(\overrightarrow{AP} = \begin{bmatrix} 1 + p \\ -5 - p \\ 1 + p \end{bmatrix}\) | M1A1 | Finding \(\overrightarrow{AP}\) |
| \(\overrightarrow{AP} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = (1 + p) - (-5 - p) + (1 + p)\) | m1 | |
| \(= 7 + 3p\) | A1 | AG; SC working with \(\lambda\) instead of \(p\) giving \(7 + 3\lambda\) \(\frac{3}{4}\) |
| (ii) \(7 + 3p = 0\) | M1 | |
| \(p = -\frac{7}{3}\), \(P\) is \(\left(-\frac{1}{3}, \frac{4}{3}, \frac{2}{3}\right)\) | A1A1 | Allow column vectors and decimals |
**(a)** $|AB| = \sqrt{(2-1)^2 + (-1-4)^2 + (3-2)^2} = \sqrt{27} = 3\sqrt{3}$ | M1, A1 | AG; convincingly obtained
**(b)** $\begin{bmatrix} -5 \\ \square \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = 1 + 5 + 1$ | M1 | Their $\overrightarrow{AB} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$
$7 = 3\sqrt{3}\sqrt{3}\cos\theta$, $\cos\theta = \frac{7}{9}$ | M1, A1 | Their scalar product $= |-1|3\sqrt{3}\cos\theta$; AG; convincingly obtained
**(c)(i)** $\overrightarrow{OP} = \begin{bmatrix} 2 + p \\ -1 - p \\ 3 + p \end{bmatrix}$, $\overrightarrow{AP} = \begin{bmatrix} 1 + p \\ -5 - p \\ 1 + p \end{bmatrix}$ | M1A1 | Finding $\overrightarrow{AP}$
$\overrightarrow{AP} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = (1 + p) - (-5 - p) + (1 + p)$ | m1 |
$= 7 + 3p$ | A1 | AG; SC working with $\lambda$ instead of $p$ giving $7 + 3\lambda$ $\frac{3}{4}$
**(ii)** $7 + 3p = 0$ | M1 |
$p = -\frac{7}{3}$, $P$ is $\left(-\frac{1}{3}, \frac{4}{3}, \frac{2}{3}\right)$ | A1A1 | Allow column vectors and decimals
**Total: 12 marks**
---7 The points $A$ and $B$ have coordinates $( 1,4,2 )$ and $( 2 , - 1,3 )$ respectively.\\
The line $l$ has equation $\mathbf { r } = \left[ \begin{array} { r } 2 \\ - 1 \\ 3 \end{array} \right] + \lambda \left[ \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance between the points $A$ and $B$ is $3 \sqrt { 3 }$.
\item The line $A B$ makes an acute angle $\theta$ with $l$. Show that $\cos \theta = \frac { 7 } { 9 }$.
\item The point $P$ on the line $l$ is where $\lambda = p$.
\begin{enumerate}[label=(\roman*)]
\item Show that
$$\overrightarrow { A P } \cdot \left[ \begin{array} { r }
1 \\
- 1 \\
1
\end{array} \right] = 7 + 3 p$$
\item Hence find the coordinates of the foot of the perpendicular from the point $A$ to the line $l$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2005 Q7 [12]}}