| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of triangle after finding foot of perpendicular or intersection |
| Difficulty | Challenging +1.2 This is a multi-part 3D vectors question requiring standard techniques: finding a line equation from two points, using perpendicularity conditions with a parameter, and calculating triangle area using the cross product. Part (b) involves solving a quadratic equation from the dot product condition AC·BC=0, which is more demanding than routine exercises but follows a predictable method. The 'show that' format and multi-step nature elevate it above average difficulty, but it remains a standard C4/FP4 examination question without requiring novel geometric insight. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\overrightarrow{AB} = (7\mathbf{i} - \mathbf{j} + 12\mathbf{k}) - (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) = (10\mathbf{i} - 4\mathbf{j} + 10\mathbf{k})\) | M1 | |
| \(\therefore \mathbf{r} = (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) + \lambda(5\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})\) | A1 | |
| (b) \(\overrightarrow{OC} = [\mu\mathbf{i} + (5-2\mu)\mathbf{j} + (-7+7\mu)\mathbf{k}]\) | M1 | |
| \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(3+\mu)\mathbf{i} + (2-2\mu)\mathbf{j} + (-9+7\mu)\mathbf{k}]\) | M1 A1 | |
| \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = [(-7+\mu)\mathbf{i} + (6-2\mu)\mathbf{j} + (-19+7\mu)\mathbf{k}]\) | A1 | |
| \(\overrightarrow{AC} \cdot \overrightarrow{BC} = (3+\mu)(-7+\mu) + (2-2\mu)(6-2\mu) + (-9+7\mu)(-19+7\mu) = 0\) | M1 | |
| \(\mu^2 - 4\mu + 3 = 0\) | A1 | |
| \((\mu-1)(\mu-3) = 0\) | M1 | |
| \(\mu = 1, 3 \quad \therefore \overrightarrow{OC} = (\mathbf{i} + 3\mathbf{j})\) or \((3\mathbf{i} - \mathbf{j} + 14\mathbf{k})\) | A2 | |
| (c) \(AC = \sqrt{16+0+4} = 2\sqrt{5}, \quad BC = \sqrt{36+16+144} = 14\) | M1 | |
| area \(= \frac{1}{2} \times 2\sqrt{5} \times 14 = 14\sqrt{5}\) | M1 A1 | (13) |
| Answer | Marks |
|---|---|
| Total | (75) |
**(a)** $\overrightarrow{AB} = (7\mathbf{i} - \mathbf{j} + 12\mathbf{k}) - (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) = (10\mathbf{i} - 4\mathbf{j} + 10\mathbf{k})$ | M1 |
$\therefore \mathbf{r} = (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) + \lambda(5\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})$ | A1 |
**(b)** $\overrightarrow{OC} = [\mu\mathbf{i} + (5-2\mu)\mathbf{j} + (-7+7\mu)\mathbf{k}]$ | M1 |
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(3+\mu)\mathbf{i} + (2-2\mu)\mathbf{j} + (-9+7\mu)\mathbf{k}]$ | M1 A1 |
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = [(-7+\mu)\mathbf{i} + (6-2\mu)\mathbf{j} + (-19+7\mu)\mathbf{k}]$ | A1 |
$\overrightarrow{AC} \cdot \overrightarrow{BC} = (3+\mu)(-7+\mu) + (2-2\mu)(6-2\mu) + (-9+7\mu)(-19+7\mu) = 0$ | M1 |
$\mu^2 - 4\mu + 3 = 0$ | A1 |
$(\mu-1)(\mu-3) = 0$ | M1 |
$\mu = 1, 3 \quad \therefore \overrightarrow{OC} = (\mathbf{i} + 3\mathbf{j})$ or $(3\mathbf{i} - \mathbf{j} + 14\mathbf{k})$ | A2 |
**(c)** $AC = \sqrt{16+0+4} = 2\sqrt{5}, \quad BC = \sqrt{36+16+144} = 14$ | M1 |
area $= \frac{1}{2} \times 2\sqrt{5} \times 14 = 14\sqrt{5}$ | M1 A1 | **(13)**
---
**Total** | **(75)**8. The line $l _ { 1 }$ passes through the points $A$ and $B$ with position vectors $( - 3 \mathbf { i } + 3 \mathbf { j } + 2 \mathbf { k } )$ and ( $7 \mathbf { i } - \mathbf { j } + 12 \mathbf { k }$ ) respectively, relative to a fixed origin.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for $l _ { 1 }$.
The line $l _ { 2 }$ has the equation
$$\mathbf { r } = ( 5 \mathbf { j } - 7 \mathbf { k } ) + \mu ( \mathbf { i } - 2 \mathbf { j } + 7 \mathbf { k } )$$
The point $C$ lies on $l _ { 2 }$ and is such that $A C$ is perpendicular to $B C$.
\item Show that one possible position vector for $C$ is $( \mathbf { i } + 3 \mathbf { j } )$ and find the other.
Assuming that $C$ has position vector $( \mathbf { i } + 3 \mathbf { j } )$,
\item find the area of triangle $A B C$, giving your answer in the form $k \sqrt { 5 }$.\\
8. continued\\
8. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q8 [13]}}