| Exam Board | OCR |
|---|---|
| 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 | Standard +0.8 This is a substantial multi-part 3D vectors question requiring: (i) standard line equation from two points, (ii) finding a point on a line satisfying a perpendicularity condition (requires setting up and solving a quadratic equation), and (iii) triangle area using cross product. Part (ii) involves non-trivial algebraic manipulation and geometric insight about perpendicular vectors, elevating this above routine C4 exercises. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB} = (7\mathbf{i} - \mathbf{j} + 12\mathbf{k}) - (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) = (10\mathbf{i} - 4\mathbf{j} + 10\mathbf{k})\) | M1 | |
| \(\therefore \mathbf{r} = (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) + \lambda(5\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OC} = [\mu\mathbf{i} + (5 - 2\mu)\mathbf{j} + (-7 + 7\mu)\mathbf{k}]\) | ||
| \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(3 + \mu)\mathbf{i} + (2 - 2\mu)\mathbf{j} + (-9 + 7\mu)\mathbf{k}]\) | 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 | |
| \(\mu = 1, 3\) \(\therefore \overrightarrow{OC} = (\mathbf{i} + 3\mathbf{j})\) or \((3\mathbf{i} - \mathbf{j} + 14\mathbf{k})\) | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AC = \sqrt{16 + 0 + 4} = 2\sqrt{5}\), \(BC = \sqrt{36 + 16 + 144} = 14\) | M1 | |
| area \(= \frac{1}{2} \times 2\sqrt{5} \times 14 = 14\sqrt{5}\) | M1 A1 | (13) |
# Question 8:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = (7\mathbf{i} - \mathbf{j} + 12\mathbf{k}) - (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) = (10\mathbf{i} - 4\mathbf{j} + 10\mathbf{k})$ | M1 | |
| $\therefore \mathbf{r} = (-3\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) + \lambda(5\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})$ | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OC} = [\mu\mathbf{i} + (5 - 2\mu)\mathbf{j} + (-7 + 7\mu)\mathbf{k}]$ | | |
| $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = [(3 + \mu)\mathbf{i} + (2 - 2\mu)\mathbf{j} + (-9 + 7\mu)\mathbf{k}]$ | 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 | |
| $\mu = 1, 3$ $\therefore \overrightarrow{OC} = (\mathbf{i} + 3\mathbf{j})$ or $(3\mathbf{i} - \mathbf{j} + 14\mathbf{k})$ | A2 | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AC = \sqrt{16 + 0 + 4} = 2\sqrt{5}$, $BC = \sqrt{36 + 16 + 144} = 14$ | M1 | |
| area $= \frac{1}{2} \times 2\sqrt{5} \times 14 = 14\sqrt{5}$ | M1 A1 | **(13)** |
---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.\\
(i) 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 $B C$.\\
(ii) Show that one possible position vector for $C$ is $( \mathbf { i } + 3 \mathbf { j } )$ and find the other.
Assuming that $C$ has position vector $( \mathbf { i } + 3 \mathbf { j } )$,\\
(iii) find the area of triangle $A B C$, giving your answer in the form $k \sqrt { 5 }$.\\
\hfill \mbox{\textit{OCR C4 Q8 [13]}}