Challenging +1.2 This is a standard Further Maths question on skew lines requiring the cross product to find a perpendicular vector and the scalar triple product formula for distance. While it involves multiple steps and Further Maths content (making it harder than average A-level), the method is direct and well-practiced with no novel insight required.
4 Two skew lines have equations \(\mathbf { r } = \left( \begin{array} { r } - 4 \\ 2 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { r } 2 \\ 0 \\ - 1 \end{array} \right)\) and \(\mathbf { r } = \left( \begin{array} { l } 6 \\ 5 \\ 2 \end{array} \right) + \mu \left( \begin{array} { l } 5 \\ 8 \\ 3 \end{array} \right)\). Find a vector which is perpendicular to both lines and determine the shortest distance between the two lines.
\((10\mathbf{i} + 3\mathbf{j} + \mathbf{k}) \cdot (8\mathbf{i} - 11\mathbf{j} + 16\mathbf{k}) = 63\) ft scalar product of their \((\mathbf{c}-\mathbf{a})\) and their \(\mathbf{n}\): B1
Shortest distance \(= 3\) cao: A1
Total: 6 marks
*EITHER:* $\mathbf{b} \times \mathbf{d} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & -1 \\ 5 & 8 & 3\end{vmatrix}$ for attempt at cross product of the two d.v.s: M1
*OR:* $\begin{pmatrix}1\\a\\b\end{pmatrix} \cdot \begin{pmatrix}2\\0\\-1\end{pmatrix} = 0$, $\begin{pmatrix}1\\a\\b\end{pmatrix} \cdot \begin{pmatrix}5\\8\\3\end{pmatrix} = 0$ and solving simultaneously for $a, b$: M1
$8\mathbf{i} - 11\mathbf{j} + 16\mathbf{k}$: A1
Shortest distance $= |(\mathbf{c}-\mathbf{a}) \cdot \hat{\mathbf{n}}|$, for use of this: M1
$|\mathbf{n}| = 21$ **ft**: B1
$\mathbf{c} - \mathbf{a} = 10\mathbf{i} + 3\mathbf{j} + \mathbf{k}$
$(10\mathbf{i} + 3\mathbf{j} + \mathbf{k}) \cdot (8\mathbf{i} - 11\mathbf{j} + 16\mathbf{k}) = 63$ **ft** scalar product of their $(\mathbf{c}-\mathbf{a})$ and their $\mathbf{n}$: B1
Shortest distance $= 3$ **cao**: A1
**Total: 6 marks**
4 Two skew lines have equations $\mathbf { r } = \left( \begin{array} { r } - 4 \\ 2 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { r } 2 \\ 0 \\ - 1 \end{array} \right)$ and $\mathbf { r } = \left( \begin{array} { l } 6 \\ 5 \\ 2 \end{array} \right) + \mu \left( \begin{array} { l } 5 \\ 8 \\ 3 \end{array} \right)$. Find a vector which is perpendicular to both lines and determine the shortest distance between the two lines.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q4 [6]}}