| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.2 This is a structured multi-part Further Maths question on 3D vectors requiring standard techniques: point-to-plane distance (formula application), finding line equations from plane intersections, point-to-line distance (cross product method), and skew line distance. Part (v) adds mild challenge using volume formula. While lengthy (5 parts), each step follows established procedures without requiring novel geometric insight—moderately above average difficulty due to Further Maths content and computational length. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Eliminating \(x\): \(3y + 4z = 9\), \(x = 7 - \frac{3}{4}t\), \(y = 3 - \frac{3}{4}t\), \(z = t\) | M1 A1A1 | Eliminating one of \(x, y, z\) or \(3x + 5z = 21\) or \(4x - 5y = 13\) (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix} = 0\) | M1 A1 ft | Appropriate scalar product |
| \(25\lambda - 25 + 16\lambda - 24 + 9\lambda - 6 = 0\) | A1 ft | |
| \(\lambda = 1.1\), \(F\) is \((7.5, 3.4, -0.3)\) | M1 | Obtaining a value for \(\lambda\) |
| \(CF = \sqrt{(0.5)^2 + (-1.6)^2 + (-1.3)^2} = \sqrt{4.5}\) | M1 A1 | Finding magnitude (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} -2 \\ -7 \\ 7 \end{pmatrix} + \mu \begin{pmatrix} 9 \\ 12 \\ -6 \end{pmatrix} - \lambda \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} - \mu \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix}\) | M1 A1 ft | Two appropriate scalar products |
| Two equations: \(-5\lambda + 9\mu - 4 = 0\) and \(-4\lambda + 12\mu - 6 = 0\) and \(3\lambda - 6\mu + 4 = 0\) | A1 ft | |
| \(\lambda = \frac{4}{81}\), \(\mu = \frac{128}{243}\) | M1 | Obtaining values for \(\lambda\) and \(\mu\) |
| Distance is \(\sqrt{\left(\frac{40}{81}\right)^2 + \left(\frac{10}{81}\right)^2 + \left(\frac{80}{81}\right)^2} = \frac{10}{9}\) | M1 A1 | Obtaining distance (6 marks) |
### 1(ii)
Eliminating $x$: $3y + 4z = 9$, $x = 7 - \frac{3}{4}t$, $y = 3 - \frac{3}{4}t$, $z = t$ | M1 A1A1 | Eliminating one of $x, y, z$ or $3x + 5z = 21$ or $4x - 5y = 13$ (3 marks)
### 1(iii)
$\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix} = 0$ | M1 A1 ft | Appropriate scalar product
$25\lambda - 25 + 16\lambda - 24 + 9\lambda - 6 = 0$ | A1 ft |
$\lambda = 1.1$, $F$ is $(7.5, 3.4, -0.3)$ | M1 | Obtaining a value for $\lambda$
$CF = \sqrt{(0.5)^2 + (-1.6)^2 + (-1.3)^2} = \sqrt{4.5}$ | M1 A1 | Finding magnitude (6 marks)
### 1(iv)
$\begin{pmatrix} -2 \\ -7 \\ 7 \end{pmatrix} + \mu \begin{pmatrix} 9 \\ 12 \\ -6 \end{pmatrix} - \lambda \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} - \mu \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix}$ | M1 A1 ft | Two appropriate scalar products
Two equations: $-5\lambda + 9\mu - 4 = 0$ and $-4\lambda + 12\mu - 6 = 0$ and $3\lambda - 6\mu + 4 = 0$ | A1 ft |
$\lambda = \frac{4}{81}$, $\mu = \frac{128}{243}$ | M1 | Obtaining values for $\lambda$ and $\mu$
Distance is $\sqrt{\left(\frac{40}{81}\right)^2 + \left(\frac{10}{81}\right)^2 + \left(\frac{80}{81}\right)^2} = \frac{10}{9}$ | M1 A1 | Obtaining distance (6 marks)1 The points $\mathrm { A } ( 2 , - 1,3 ) , \mathrm { B } ( - 2 , - 7,7 )$ and $\mathrm { C } ( 7,5,1 )$ are three vertices of a tetrahedron ABCD .\\
The plane ABD has equation $x + 4 y + 7 z = 19$.\\
The plane ACD has equation $2 x - y + 2 z = 11$.\\
(i) Find the shortest distance from $B$ to the plane $A C D$.\\
(ii) Find an equation for the line AD .\\
(iii) Find the shortest distance from C to the line AD .\\
(iv) Find the shortest distance between the lines $A D$ and $B C$.\\
(v) Given that the tetrahedron ABCD has volume 20, find the coordinates of the two possible positions for the vertex $D$.
\hfill \mbox{\textit{OCR MEI FP3 2011 Q1 [24]}}