| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | June |
| Marks | 2 |
| 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 cross product calculation and application of the distance formula. While it involves multiple steps (finding perpendicular vector via cross product, then using scalar triple product for distance), these are well-rehearsed techniques for FM students with no novel insight required. Slightly above average due to being FM content and multi-step, but routine for this level. |
| Spec | 4.04g Vector product: a x b perpendicular vector4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) Sh. Dist. \(= | (\mathbf{b}-\mathbf{a})\cdot\hat{\mathbf{n}} | \), \((\mathbf{b}-\mathbf{a}) = \pm(\mathbf{i}+3\mathbf{j}-7\mathbf{k})\), \(\hat{\mathbf{n}} = \frac{1}{\sqrt{38}}(2\mathbf{i}+5\mathbf{j}-3\mathbf{k})\) ft M1 B1 B1 |
| \(= \frac{1}{\sqrt{38}}(2\mathbf{i}+5\mathbf{j}-3\mathbf{k})\cdot(\mathbf{i}+3\mathbf{j}-7\mathbf{k}) = \frac{1}{\sqrt{38}} | 2+15+21 | \) ft scalar prod. B1 |
**(i)** $\mathbf{d}_1 \times \mathbf{d}_2$ attempted **M1**; $= 14\mathbf{i} + 35\mathbf{j} - 21\mathbf{k}$ **A1**
(ALT: Use of 2 scalar products & attempt to get 2 components in terms of the 3rd)
**[2]**
**(ii)** Sh. Dist. $= |(\mathbf{b}-\mathbf{a})\cdot\hat{\mathbf{n}}|$, $(\mathbf{b}-\mathbf{a}) = \pm(\mathbf{i}+3\mathbf{j}-7\mathbf{k})$, $\hat{\mathbf{n}} = \frac{1}{\sqrt{38}}(2\mathbf{i}+5\mathbf{j}-3\mathbf{k})$ **ft** **M1 B1 B1**
$= \frac{1}{\sqrt{38}}(2\mathbf{i}+5\mathbf{j}-3\mathbf{k})\cdot(\mathbf{i}+3\mathbf{j}-7\mathbf{k}) = \frac{1}{\sqrt{38}}|2+15+21|$ **ft** scalar prod. **B1**
$= \sqrt{38}$ **cao** **A1**
**ALT:** Solving $\begin{pmatrix}2\\4\\3\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\4\end{pmatrix}-\mu\begin{pmatrix}1\\1\\10\end{pmatrix} - \begin{pmatrix}9\\-3\\1\end{pmatrix} = k\begin{pmatrix}2\\5\\-3\end{pmatrix}$ to find closest points on line, $(3,6,7)$ from $\lambda=1$ and $(1,1,10)$ from $\mu=0$ giving $k=1$ and Sh.D. $= \sqrt{38}$
**[5]**4 (i) Find a vector which is perpendicular to both of the vectors
$$\mathbf { d } _ { 1 } = \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } \quad \text { and } \quad \mathbf { d } _ { 2 } = 9 \mathbf { i } - 3 \mathbf { j } + \mathbf { k } .$$
(ii) Determine the shortest distance between the skew lines with equations
$$\mathbf { r } = 2 \mathbf { i } + 4 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = \mathbf { i } + \mathbf { j } + 10 \mathbf { k } + \mu ( 9 \mathbf { i } - 3 \mathbf { j } + \mathbf { k } ) .$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q4 [2]}}