| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring multiple vector techniques: finding shortest distance between skew lines using the scalar triple product formula, finding point-to-line distance, determining when lines intersect (requiring solving simultaneous equations), and calculating tetrahedron volume. While the individual techniques are standard FP3 content, the multi-part nature, computational complexity, and need to correctly apply several different formulas makes this moderately challenging even for Further Maths students. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} \times \overrightarrow{BC} = \begin{pmatrix}8\\-2\\-13\end{pmatrix} \times \begin{pmatrix}-6\\18\\3\end{pmatrix} = \begin{pmatrix}228\\54\\132\end{pmatrix}\) | M1*, A2 | Appropriate vector product; Give A1 if one error |
| \(\ | \overrightarrow{AB} \times \overrightarrow{BC}\ | = \sqrt{228^2 + 54^2 + 132^2}\); \(\ |
| Shortest distance is \(\dfrac{\ | \overrightarrow{AB} \times \overrightarrow{BC}\ | }{\ |
| Shortest distance is \(14\) | A1 [6] | Sign error in vector product can earn M1A1M1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\begin{pmatrix}11-6\lambda\\18\lambda\\-3+3\lambda\end{pmatrix} - \begin{pmatrix}3\\2\\10\end{pmatrix}\right] \cdot \begin{pmatrix}-6\\18\\3\end{pmatrix} = 0\) | M1*, A1 | Allow one error |
| \(\lambda = \dfrac{1}{3}\) | M1*, A1 | Dep*; Obtaining a value of \(\lambda\) |
| Shortest distance is \(\sqrt{(6)^2+(4)^2+(-12)^2}\) | M1 | Dep** |
| Shortest distance is \(14\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}11\\0\\-3\end{pmatrix} + \lambda\begin{pmatrix}-6\\18\\k+3\end{pmatrix} = \begin{pmatrix}3\\2\\10\end{pmatrix} + \mu\begin{pmatrix}-1\\4\\1\end{pmatrix}\) | ||
| \(11-6\lambda = 3-\mu\) | M1, A1 | Allow one error; Two correct equations; Must use different parameters |
| \(18\lambda = 2+4\mu\) | ||
| \(\lambda=5,\; \mu=22\) | A1 | |
| \(-3+\lambda(k+3) = 10+\mu\) | M1 | Obtaining a value of \(k\) |
| \(k=4\) | A1 | Other methods possible (e.g. distance between lines is 0) |
| Point of intersection is \(\begin{pmatrix}3\\2\\10\end{pmatrix} + 22\begin{pmatrix}-1\\4\\1\end{pmatrix}\) | M1 | |
| Point of intersection is \((-19,\; 90,\; 32)\) | A1 [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left\ | \begin{pmatrix}-1\\4\\1\end{pmatrix}\right\ | = \sqrt{18}\), so \(\overrightarrow{AD} = (\pm)\dfrac{12}{\sqrt{18}}\begin{pmatrix}-1\\4\\1\end{pmatrix} = 2\sqrt{2}\begin{pmatrix}-1\\4\\1\end{pmatrix}\) |
| Volume is \(\dfrac{1}{6}(\overrightarrow{AB} \times \overrightarrow{AC})\cdot\overrightarrow{AD}\) | M1* | Appropriate scalar triple product |
| \(= \dfrac{1}{6}\left[\begin{pmatrix}8\\-2\\-13\end{pmatrix} \times \begin{pmatrix}2\\16\\-10\end{pmatrix}\right] \cdot (2\sqrt{2})\begin{pmatrix}-1\\4\\1\end{pmatrix}\) | A1 ft | Correct expression; Can be implied |
| \(= \dfrac{\sqrt{2}}{3}\begin{pmatrix}228\\54\\132\end{pmatrix}\cdot\begin{pmatrix}-1\\4\\1\end{pmatrix} = \dfrac{\sqrt{2}}{3}(120)\) | M1 | Evaluating scalar triple product; Dep** |
| \(= 40\sqrt{2}\) | A1 [6] | Accept 56.6 |
## Question 1:
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} \times \overrightarrow{BC} = \begin{pmatrix}8\\-2\\-13\end{pmatrix} \times \begin{pmatrix}-6\\18\\3\end{pmatrix} = \begin{pmatrix}228\\54\\132\end{pmatrix}$ | M1*, A2 | Appropriate vector product; Give A1 if one error |
| $\|\overrightarrow{AB} \times \overrightarrow{BC}\| = \sqrt{228^2 + 54^2 + 132^2}$; $\|\overrightarrow{BC}\| = \sqrt{6^2+18^2+3^2}$ | M1* | Dep* |
| Shortest distance is $\dfrac{\|\overrightarrow{AB} \times \overrightarrow{BC}\|}{\|\overrightarrow{BC}\|} = \sqrt{\dfrac{72324}{369}}$ | M1 | Dep** |
| Shortest distance is $14$ | A1 [6] | Sign error in vector product can earn M1A1M1M1A1 |
**OR method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\begin{pmatrix}11-6\lambda\\18\lambda\\-3+3\lambda\end{pmatrix} - \begin{pmatrix}3\\2\\10\end{pmatrix}\right] \cdot \begin{pmatrix}-6\\18\\3\end{pmatrix} = 0$ | M1*, A1 | Allow one error |
| $\lambda = \dfrac{1}{3}$ | M1*, A1 | Dep*; Obtaining a value of $\lambda$ |
| Shortest distance is $\sqrt{(6)^2+(4)^2+(-12)^2}$ | M1 | Dep** |
| Shortest distance is $14$ | A1 | |
---
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}11\\0\\-3\end{pmatrix} + \lambda\begin{pmatrix}-6\\18\\k+3\end{pmatrix} = \begin{pmatrix}3\\2\\10\end{pmatrix} + \mu\begin{pmatrix}-1\\4\\1\end{pmatrix}$ | | |
| $11-6\lambda = 3-\mu$ | M1, A1 | Allow one error; Two correct equations; Must use different parameters |
| $18\lambda = 2+4\mu$ | | |
| $\lambda=5,\; \mu=22$ | A1 | |
| $-3+\lambda(k+3) = 10+\mu$ | M1 | Obtaining a value of $k$ |
| $k=4$ | A1 | Other methods possible (e.g. distance between lines is 0) |
| Point of intersection is $\begin{pmatrix}3\\2\\10\end{pmatrix} + 22\begin{pmatrix}-1\\4\\1\end{pmatrix}$ | M1 | |
| Point of intersection is $(-19,\; 90,\; 32)$ | A1 [7] | |
---
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left\|\begin{pmatrix}-1\\4\\1\end{pmatrix}\right\| = \sqrt{18}$, so $\overrightarrow{AD} = (\pm)\dfrac{12}{\sqrt{18}}\begin{pmatrix}-1\\4\\1\end{pmatrix} = 2\sqrt{2}\begin{pmatrix}-1\\4\\1\end{pmatrix}$ | M1*, A1 | Obtaining $\overrightarrow{AD}$ or D |
| Volume is $\dfrac{1}{6}(\overrightarrow{AB} \times \overrightarrow{AC})\cdot\overrightarrow{AD}$ | M1* | Appropriate scalar triple product |
| $= \dfrac{1}{6}\left[\begin{pmatrix}8\\-2\\-13\end{pmatrix} \times \begin{pmatrix}2\\16\\-10\end{pmatrix}\right] \cdot (2\sqrt{2})\begin{pmatrix}-1\\4\\1\end{pmatrix}$ | A1 ft | Correct expression; Can be implied |
| $= \dfrac{\sqrt{2}}{3}\begin{pmatrix}228\\54\\132\end{pmatrix}\cdot\begin{pmatrix}-1\\4\\1\end{pmatrix} = \dfrac{\sqrt{2}}{3}(120)$ | M1 | Evaluating scalar triple product; Dep** |
| $= 40\sqrt{2}$ | A1 [6] | Accept 56.6 |
---1 Three points have coordinates $\mathrm { A } ( 3,2,10 ) , \mathrm { B } ( 11,0 , - 3 ) , \mathrm { C } ( 5,18,0 )$, and $L$ is the straight line through A with equation
$$\frac { x - 3 } { - 1 } = \frac { y - 2 } { 4 } = \frac { z - 10 } { 1 }$$
(i) Find the shortest distance between the lines $L$ and BC .\\
(ii) Find the shortest distance from A to the line BC .
A straight line passes through B and the point $\mathrm { P } ( 5,18 , k )$, and intersects the line $L$.\\
(iii) Find $k$, and the point of intersection of the lines BP and $L$.
The point D is on the line $L$, and AD has length 12 .\\
(iv) Find the volume of the tetrahedron ABCD .
\hfill \mbox{\textit{OCR MEI FP3 2013 Q1 [24]}}