| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from origin to line |
| Difficulty | Standard +0.8 This is a multi-part 3D vectors question requiring line equations, perpendicular distance concepts, and area calculation using the cross product. While parts (i)-(ii) are routine, parts (iii)-(iv) require understanding of perpendicularity conditions and vector area formulas, making this moderately challenging but still within standard C4 scope. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks |
|---|---|
| (i) \(\overrightarrow{AB} = \begin{pmatrix}10\\-15\\5\end{pmatrix}\); \(\therefore \mathbf{r} = \begin{pmatrix}3\\9\\-7\end{pmatrix} + \lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}\) | M1 A1 |
| (ii) \(3 + 2\lambda = 9 \therefore \lambda = 3\) | M1 |
| When \(\lambda = 3\): \(\mathbf{r} = \begin{pmatrix}9\\0\\-4\end{pmatrix}\); \(\therefore (9, 0, -4)\) lies on \(l\) | A1 |
| (iii) \(\overrightarrow{OD} = \begin{pmatrix}3 + 2\mu\\9 - 3\mu\\-7 + \mu\end{pmatrix}\); \(\therefore \begin{pmatrix}3 + 2\mu\\9 - 3\mu\\-7 + \mu\end{pmatrix} \cdot \begin{pmatrix}2\\-3\\1\end{pmatrix} = 0\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = 2\); \(\overrightarrow{OD} = \begin{pmatrix}7\\3\\-5\end{pmatrix}\); \(D(7, 3, -5)\) | M1 A1 | |
| (iv) \(AB = \sqrt{100 + 225 + 25} = \sqrt{350}\); \(OD = \sqrt{49 + 9 + 25} = \sqrt{83}\) | M1 | |
| Area \(= \frac{1}{2} \times \sqrt{350} \times \sqrt{83} = 85.2\) (3sf) | M1 A1 | (10) |
**(i)** $\overrightarrow{AB} = \begin{pmatrix}10\\-15\\5\end{pmatrix}$; $\therefore \mathbf{r} = \begin{pmatrix}3\\9\\-7\end{pmatrix} + \lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}$ | M1 A1 |
**(ii)** $3 + 2\lambda = 9 \therefore \lambda = 3$ | M1 |
When $\lambda = 3$: $\mathbf{r} = \begin{pmatrix}9\\0\\-4\end{pmatrix}$; $\therefore (9, 0, -4)$ lies on $l$ | A1 |
**(iii)** $\overrightarrow{OD} = \begin{pmatrix}3 + 2\mu\\9 - 3\mu\\-7 + \mu\end{pmatrix}$; $\therefore \begin{pmatrix}3 + 2\mu\\9 - 3\mu\\-7 + \mu\end{pmatrix} \cdot \begin{pmatrix}2\\-3\\1\end{pmatrix} = 0$ | M1 |
$6 + 4\mu - 27 + 9\mu - 7 + \mu = 0$
$\lambda = 2$; $\overrightarrow{OD} = \begin{pmatrix}7\\3\\-5\end{pmatrix}$; $D(7, 3, -5)$ | M1 A1 |
**(iv)** $AB = \sqrt{100 + 225 + 25} = \sqrt{350}$; $OD = \sqrt{49 + 9 + 25} = \sqrt{83}$ | M1 |
Area $= \frac{1}{2} \times \sqrt{350} \times \sqrt{83} = 85.2$ (3sf) | M1 A1 | **(10)**
---8. The points $A$ and $B$ have coordinates $( 3,9 , - 7 )$ and $( 13 , - 6 , - 2 )$ respectively.\\
(i) Find, in vector form, an equation for the line $l$ which passes through $A$ and $B$.\\
(ii) Show that the point $C$ with coordinates $( 9,0 , - 4 )$ lies on $l$.
The point $D$ is the point on $l$ closest to the origin, $O$.\\
(iii) Find the coordinates of $D$.\\
(iv) Find the area of triangle $O A B$ to 3 significant figures.\\
\hfill \mbox{\textit{OCR C4 Q8 [10]}}