2 Find a vector which is perpendicular to both of the lines
$$\mathbf { r } = \left( \begin{array} { r }
11 \\
5 \\
4
\end{array} \right) + \lambda \left( \begin{array} { l }
6 \\
2 \\
5
\end{array} \right) \quad \text { and } \quad \mathbf { r } = \left( \begin{array} { r }
1 \\
7 \\
- 1
\end{array} \right) + \mu \left( \begin{array} { r }
- 6 \\
1 \\
4
\end{array} \right)$$
and hence find the shortest distance between them.
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Question 2 [6 marks]\(\begin{pmatrix}6\\2\\5\end{pmatrix} \times \begin{pmatrix}-6\\1\\4\end{pmatrix} = 3\begin{pmatrix}1\\-18\\6\end{pmatrix}\) M1 (Attempt at vector products of the d.v.s), A1 (any suitable multiple)
Answer Marks
Guidance
Shortest Distance \(= (\mathbf{b}-\mathbf{a}) \bullet \hat{\mathbf{n}}
\) M1
\(= \frac{1}{19}\begin{pmatrix}10\\-2\\5\end{pmatrix} \bullet \begin{pmatrix}1\\-18\\6\end{pmatrix} = \frac{1}{19}(10+36+30)\) B1 (\( \hat{\mathbf{n}}
\) correct), B1 (Sc. Prod. ft correct)
\(= 4\) A1
[6] Alternative method:
M1 A1 for common normal \(\mathbf{i} - 18\mathbf{j} + 6\mathbf{k}\)M1 A1 for parallel planes \(x - 18y + 6z = -55\) and \(-131\)
Answer Marks
Guidance
M1 A1 for Sh.D formula, \(\frac{131-55
}{
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**Question 2** [6 marks]
$\begin{pmatrix}6\\2\\5\end{pmatrix} \times \begin{pmatrix}-6\\1\\4\end{pmatrix} = 3\begin{pmatrix}1\\-18\\6\end{pmatrix}$ M1 (Attempt at vector products of the d.v.s), A1 (any suitable multiple)
Shortest Distance $= |(\mathbf{b}-\mathbf{a}) \bullet \hat{\mathbf{n}}|$ M1
$= \frac{1}{19}\begin{pmatrix}10\\-2\\5\end{pmatrix} \bullet \begin{pmatrix}1\\-18\\6\end{pmatrix} = \frac{1}{19}(10+36+30)$ B1 ($|\hat{\mathbf{n}}|$ correct), B1 (Sc. Prod. **ft** correct)
$= 4$ A1
**[6]**
Alternative method:
**M1 A1** for common normal $\mathbf{i} - 18\mathbf{j} + 6\mathbf{k}$
**M1 A1** for parallel planes $x - 18y + 6z = -55$ and $-131$
**M1 A1** for Sh.D formula, $\frac{|131-55|}{|\mathbf{n}|} = \frac{76}{19} = 4$
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2 Find a vector which is perpendicular to both of the lines
$$\mathbf { r } = \left( \begin{array} { r }
11 \\
5 \\
4
\end{array} \right) + \lambda \left( \begin{array} { l }
6 \\
2 \\
5
\end{array} \right) \quad \text { and } \quad \mathbf { r } = \left( \begin{array} { r }
1 \\
7 \\
- 1
\end{array} \right) + \mu \left( \begin{array} { r }
- 6 \\
1 \\
4
\end{array} \right)$$
and hence find the shortest distance between them.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q2 [6]}}