| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| 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 | Challenging +1.2 Part (a) is a standard angle-between-lines calculation using dot product. Part (b) requires finding a point on a line satisfying a perpendicularity condition—straightforward but involves setting up and solving BC·BA = 0. Part (c) is more involved, requiring construction of point D using parallel and perpendicular conditions, then finding E on line BD with a distance constraint, yielding two solutions. While multi-step, each component uses routine vector techniques without requiring novel insight. Slightly above average due to the length and coordination required in part (c). |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Direction of \(l\): \(\begin{pmatrix}3\\1\\-2\end{pmatrix}\), Direction of \(AB\): \(\begin{pmatrix}2\\-4\\-6\end{pmatrix}\) | M1 | Finding vector \(\overrightarrow{AB} = B - A = \begin{pmatrix}5-3\\-2-2\\4-10\end{pmatrix}\) |
| \(\cos\theta = \frac{\begin{pmatrix}3\\1\\-2\end{pmatrix}\cdot\begin{pmatrix}2\\-4\\-6\end{pmatrix}}{ | \begin{pmatrix}3\\1\\-2\end{pmatrix} | |
| Numerator: \(6 - 4 + 12 = 14\) | A1 | |
| \( | \mathbf{d}_l | = \sqrt{9+1+4} = \sqrt{14}\), \( |
| \(\cos\theta = \frac{14}{\sqrt{14}\cdot\sqrt{56}} = \frac{14}{\sqrt{784}} = \frac{14}{28} = \frac{1}{2}\) | A1 | \(\theta = 60°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(C = \begin{pmatrix}3+3\lambda\\2+\lambda\\10-2\lambda\end{pmatrix}\) for some \(\lambda\) | Point on \(l\) | |
| \(\overrightarrow{BC} = \begin{pmatrix}3+3\lambda-5\\2+\lambda+2\\10-2\lambda-4\end{pmatrix} = \begin{pmatrix}3\lambda-2\\\lambda+4\\6-2\lambda\end{pmatrix}\) | M1 | Finding \(\overrightarrow{BC}\) |
| \(\overrightarrow{BC}\cdot\overrightarrow{AB} = 0\) (since angle \(ABC = 90°\)) | M1 | Using perpendicularity condition |
| \(2(3\lambda-2) + (-4)(\lambda+4) + (-6)(6-2\lambda) = 0\) | ||
| \(6\lambda - 4 - 4\lambda - 16 - 36 + 12\lambda = 0\) | A1 | |
| \(14\lambda - 56 = 0 \Rightarrow \lambda = 4\) | ||
| \(C = (15, 6, 2)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(BD\) parallel to \(AC\), so direction of \(BD\) is direction of \(l\): \(\begin{pmatrix}3\\1\\-2\end{pmatrix}\) | M1 | |
| \(\overrightarrow{AC} = \begin{pmatrix}12\\4\\-8\end{pmatrix}\), \( | AC | = \sqrt{144+16+64} = \sqrt{224} = 4\sqrt{14}\) |
| \(D = \begin{pmatrix}5+3\mu\\-2+\mu\\4-2\mu\end{pmatrix}\) | \(B\) plus multiple of direction | |
| Angle \(BCD = 90°\), so \(\overrightarrow{CB}\cdot\overrightarrow{CD} = 0\)... or use \( | BD | = |
| \( | BD | = |
| \(\mu = 4\): \(D = (17, 2, -4)\) | A1 | |
| \(\mu = -4\): \(D = (-7, -6, 12)\) | A1 | Both positions required |
| \(E\) is midpoint of \(BD\) extended such that \(DE = \frac{1}{2}AC\) | ||
| \(E = B + \frac{3}{2}\overrightarrow{BD}\)... giving two positions |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Direction of $l$: $\begin{pmatrix}3\\1\\-2\end{pmatrix}$, Direction of $AB$: $\begin{pmatrix}2\\-4\\-6\end{pmatrix}$ | M1 | Finding vector $\overrightarrow{AB} = B - A = \begin{pmatrix}5-3\\-2-2\\4-10\end{pmatrix}$ |
| $\cos\theta = \frac{\begin{pmatrix}3\\1\\-2\end{pmatrix}\cdot\begin{pmatrix}2\\-4\\-6\end{pmatrix}}{|\begin{pmatrix}3\\1\\-2\end{pmatrix}||\begin{pmatrix}2\\-4\\-6\end{pmatrix}|}$ | M1 | Correct dot product formula used |
| Numerator: $6 - 4 + 12 = 14$ | A1 | |
| $|\mathbf{d}_l| = \sqrt{9+1+4} = \sqrt{14}$, $|\overrightarrow{AB}| = \sqrt{4+16+36} = \sqrt{56}$ | | |
| $\cos\theta = \frac{14}{\sqrt{14}\cdot\sqrt{56}} = \frac{14}{\sqrt{784}} = \frac{14}{28} = \frac{1}{2}$ | A1 | $\theta = 60°$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C = \begin{pmatrix}3+3\lambda\\2+\lambda\\10-2\lambda\end{pmatrix}$ for some $\lambda$ | | Point on $l$ |
| $\overrightarrow{BC} = \begin{pmatrix}3+3\lambda-5\\2+\lambda+2\\10-2\lambda-4\end{pmatrix} = \begin{pmatrix}3\lambda-2\\\lambda+4\\6-2\lambda\end{pmatrix}$ | M1 | Finding $\overrightarrow{BC}$ |
| $\overrightarrow{BC}\cdot\overrightarrow{AB} = 0$ (since angle $ABC = 90°$) | M1 | Using perpendicularity condition |
| $2(3\lambda-2) + (-4)(\lambda+4) + (-6)(6-2\lambda) = 0$ | | |
| $6\lambda - 4 - 4\lambda - 16 - 36 + 12\lambda = 0$ | A1 | |
| $14\lambda - 56 = 0 \Rightarrow \lambda = 4$ | | |
| $C = (15, 6, 2)$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $BD$ parallel to $AC$, so direction of $BD$ is direction of $l$: $\begin{pmatrix}3\\1\\-2\end{pmatrix}$ | M1 | |
| $\overrightarrow{AC} = \begin{pmatrix}12\\4\\-8\end{pmatrix}$, $|AC| = \sqrt{144+16+64} = \sqrt{224} = 4\sqrt{14}$ | | |
| $D = \begin{pmatrix}5+3\mu\\-2+\mu\\4-2\mu\end{pmatrix}$ | | $B$ plus multiple of direction |
| Angle $BCD = 90°$, so $\overrightarrow{CB}\cdot\overrightarrow{CD} = 0$... or use $|BD| = |AC| = 4\sqrt{14}$ since $ABDC$ is a rectangle | M1 | |
| $|BD| = |AC| = 4\sqrt{14}$, so $|\mu|\sqrt{14} = 4\sqrt{14}$, giving $\mu = \pm 4$ | A1 | |
| $\mu = 4$: $D = (17, 2, -4)$ | A1 | |
| $\mu = -4$: $D = (-7, -6, 12)$ | A1 | Both positions required |
| $E$ is midpoint of $BD$ extended such that $DE = \frac{1}{2}AC$ | | |
| $E = B + \frac{3}{2}\overrightarrow{BD}$... giving two positions | | |
---6 The points $A$ and $B$ have coordinates $( 3,2,10 )$ and $( 5 , - 2,4 )$ respectively.\\
The line $l$ passes through $A$ and has equation $\mathbf { r } = \left[ \begin{array} { r } 3 \\ 2 \\ 10 \end{array} \right] + \lambda \left[ \begin{array} { r } 3 \\ 1 \\ - 2 \end{array} \right]$.
\begin{enumerate}[label=(\alph*)]
\item Find the acute angle between $l$ and the line $A B$.
\item The point $C$ lies on $l$ such that angle $A B C$ is $90 ^ { \circ }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{fdd3905e-11f7-4b20-adfe-4c686018a221-12_360_339_762_852}
Find the coordinates of $C$.
\item The point $D$ is such that $B D$ is parallel to $A C$ and angle $B C D$ is $90 ^ { \circ }$. The point $E$ lies on the line through $B$ and $D$ and is such that the length of $D E$ is half that of $A C$. Find the coordinates of the two possible positions of $E$.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2015 Q6 [12]}}