| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| 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 straightforward multi-part 3D vectors question requiring standard techniques: finding a vector between points, calculating an angle using the dot product formula, writing a line equation, and finding a point satisfying a perpendicularity condition. All steps are routine applications of Core 4 vector methods with no novel problem-solving required, 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 |
|---|---|---|
| Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{bmatrix}3\\1\\-2\end{bmatrix} - \begin{bmatrix}2\\1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\\-1\end{bmatrix}\) | M1, A1 | \(\pm(\overrightarrow{OA}-\overrightarrow{OB})\); A0 if answer as coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\overrightarrow{OB} \cdot \overrightarrow{AB} = 3\times1 + 1\times0 + (-2)\times(-1) = 5\) | M1, A1 | Evaluate to single value |
| \(\cos\theta = \frac{\overrightarrow{OB}\cdot\overrightarrow{AB}}{ | \overrightarrow{OB} | |
| \( | \overrightarrow{OB} | = \sqrt{14}\), \( |
| \(\cos\theta = \frac{5}{\sqrt{7}\times2\sqrt{2}} = \frac{5}{2\sqrt{7}}\) | A1 | CSO; AG so need to see intermediate step e.g. \(\frac{5}{\sqrt{7}\times2\sqrt{2}}\) or \(\frac{5}{\sqrt{28}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{r} = \begin{bmatrix}6\\2\\-4\end{bmatrix} + \lambda\begin{bmatrix}1\\0\\-1\end{bmatrix}\) | M1, A1F | \(\overrightarrow{OC} + \lambda\overrightarrow{AB}\). Allow one slip; ft on \(\overrightarrow{AB}\); needs r or \(\begin{bmatrix}x\\y\\z\end{bmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\overrightarrow{OD}\cdot\overrightarrow{AB} = \begin{bmatrix}6+\lambda\\2\\-4-\lambda\end{bmatrix}\cdot\begin{bmatrix}1\\0\\-1\end{bmatrix}\) | M1 | |
| \(6 + \lambda + 4 + \lambda = 0\) | m1 | |
| \(\lambda = -5\) | A1F | ft on equation of line |
| \(D\) is \((1,2,1)\) | A1 | CAO |
# Question 8:
## Part (a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\overrightarrow{AB} = \begin{bmatrix}3\\1\\-2\end{bmatrix} - \begin{bmatrix}2\\1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\\-1\end{bmatrix}$ | M1, A1 | $\pm(\overrightarrow{OA}-\overrightarrow{OB})$; A0 if answer as coordinates | **Total: 2** |
## Part (a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\overrightarrow{OB} \cdot \overrightarrow{AB} = 3\times1 + 1\times0 + (-2)\times(-1) = 5$ | M1, A1 | Evaluate to single value |
| $\cos\theta = \frac{\overrightarrow{OB}\cdot\overrightarrow{AB}}{|\overrightarrow{OB}||\overrightarrow{AB}|}$ | M1 | Use formula for $\cos\theta$ with any 2 vectors and at least one corresponding moduli 'correct' |
| $|\overrightarrow{OB}| = \sqrt{14}$, $|\overrightarrow{AB}| = \sqrt{2}$ | | |
| $\cos\theta = \frac{5}{\sqrt{7}\times2\sqrt{2}} = \frac{5}{2\sqrt{7}}$ | A1 | CSO; AG so need to see intermediate step e.g. $\frac{5}{\sqrt{7}\times2\sqrt{2}}$ or $\frac{5}{\sqrt{28}}$ | **Total: 4** |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{r} = \begin{bmatrix}6\\2\\-4\end{bmatrix} + \lambda\begin{bmatrix}1\\0\\-1\end{bmatrix}$ | M1, A1F | $\overrightarrow{OC} + \lambda\overrightarrow{AB}$. Allow one slip; ft on $\overrightarrow{AB}$; needs **r** or $\begin{bmatrix}x\\y\\z\end{bmatrix}$ | **Total: 2** |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\overrightarrow{OD}\cdot\overrightarrow{AB} = \begin{bmatrix}6+\lambda\\2\\-4-\lambda\end{bmatrix}\cdot\begin{bmatrix}1\\0\\-1\end{bmatrix}$ | M1 | |
| $6 + \lambda + 4 + \lambda = 0$ | m1 | |
| $\lambda = -5$ | A1F | ft on equation of line |
| $D$ is $(1,2,1)$ | A1 | CAO | **Total: 4** |8 The points $A$ and $B$ have coordinates $( 2,1 , - 1 )$ and $( 3,1 , - 2 )$ respectively. The angle $O B A$ is $\theta$, where $O$ is the origin.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the vector $\overrightarrow { A B }$.
\item Show that $\cos \theta = \frac { 5 } { 2 \sqrt { 7 } }$.
\end{enumerate}\item The point $C$ is such that $\overrightarrow { O C } = 2 \overrightarrow { O B }$. The line $l$ is parallel to $\overrightarrow { A B }$ and passes through the point $C$. Find a vector equation of $l$.
\item The point $D$ lies on $l$ such that angle $O D C = 90 ^ { \circ }$. Find the coordinates of $D$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q8 [12]}}