| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| 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 standard Further Maths question on skew lines requiring the cross product formula for shortest distance and finding a plane equation, but it's methodical application of known techniques rather than requiring novel insight. The multi-step nature and Further Maths content places it above average difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}4\\-1\\-1\end{pmatrix}=\begin{pmatrix}-4\\-2\\-14\end{pmatrix}=-2\begin{pmatrix}2\\1\\7\end{pmatrix}\) | M1, A1 | Or any multiple |
| \(\begin{pmatrix}3\\0\\1\end{pmatrix}-\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}2\\-2\\0\end{pmatrix}\) | B1 | Or negative; or use of \(\mathbf{n}\cdot(\mathbf{a_1}+p\mathbf{b_1}+k\mathbf{n})=\mathbf{n}\cdot(\mathbf{a_2}+q\mathbf{b_2})\) B1 followed by attempt to calculate magnitude of \(k\mathbf{n}\) M1 |
| shortest distance \(=\frac{\begin{vmatrix}\begin{pmatrix}2\\-2\\0\end{pmatrix}\cdot\begin{pmatrix}2\\1\\7\end{pmatrix}\end{vmatrix}}{\sqrt{2^2+1^2+7^2}}=\frac{2}{\sqrt{54}}\) oe | M1, A1 | Component of their vector in their direction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2x+y+7z=\ldots\) | B1ft | ft from 4(i) only if 1st M1 mark gained |
| \(\ldots 11\) | B1 dep | If zero, then sc1 for any correct vector equation |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2\\3\\-1\end{pmatrix}\times\begin{pmatrix}4\\-1\\-1\end{pmatrix}=\begin{pmatrix}-4\\-2\\-14\end{pmatrix}=-2\begin{pmatrix}2\\1\\7\end{pmatrix}$ | M1, A1 | Or any multiple |
| $\begin{pmatrix}3\\0\\1\end{pmatrix}-\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}2\\-2\\0\end{pmatrix}$ | B1 | Or negative; or use of $\mathbf{n}\cdot(\mathbf{a_1}+p\mathbf{b_1}+k\mathbf{n})=\mathbf{n}\cdot(\mathbf{a_2}+q\mathbf{b_2})$ B1 followed by attempt to calculate magnitude of $k\mathbf{n}$ M1 |
| shortest distance $=\frac{\begin{vmatrix}\begin{pmatrix}2\\-2\\0\end{pmatrix}\cdot\begin{pmatrix}2\\1\\7\end{pmatrix}\end{vmatrix}}{\sqrt{2^2+1^2+7^2}}=\frac{2}{\sqrt{54}}$ oe | M1, A1 | Component of their vector in their direction |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x+y+7z=\ldots$ | B1ft | ft from 4(i) only if 1st M1 mark gained |
| $\ldots 11$ | B1 dep | If zero, then sc1 for any correct vector equation |
---4 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations
$$\mathbf { r } = \left( \begin{array} { l }
1 \\
2 \\
1
\end{array} \right) + \lambda \left( \begin{array} { r }
2 \\
3 \\
- 1
\end{array} \right) \text { and } \mathbf { r } = \left( \begin{array} { l }
3 \\
0 \\
1
\end{array} \right) + \mu \left( \begin{array} { r }
4 \\
- 1 \\
- 1
\end{array} \right)$$
respectively.\\
(i) Find the shortest distance between the lines.\\
(ii) Find a cartesian equation of the plane which contains $l _ { 1 }$ and which is parallel to $l _ { 2 }$.
\hfill \mbox{\textit{OCR FP3 2013 Q4 [7]}}