vectors

$\displaystyle \begin{aligned} L_1: \vec{x}(\lambda) &= \lambda \vec{a} + \vec{b} \\ L_2: \vec{x}(\mu) &= \mu \vec{c} + \vec{d} \end{aligned}$

The minimum distance between 2 lines is

$d = | (\vec{p_1} -\vec{p_2}) \cdot \hat{n} | \quad(1)$

where $\vec{p_i}$ are position vectors of points on the respective lines, and $\hat{n}$ is normal to both lines.

If $L_1$ and $L_2$ intersect, $d=0$ . Using $(1)$ and $L_1$ and $L_2$ ,

$d = | (\vec{b} -\vec{d}) \cdot \hat{n} |$ and $\hat{n} = \frac{\vec{a} \times \vec{c}}{|\vec{a} \times \vec{c}|}$

$\therefore d = | (\vec{b} -\vec{d}) \cdot \frac{\vec{a} \times \vec{c}}{|\vec{a} \times \vec{c}|} | = 0$

$\begin{aligned} \Rightarrow (\vec{b} -\vec{d}) \cdot {\vec{a} \times \vec{c}} &= 0 \\ \Rightarrow \vec{b} \cdot \vec{a} \times \vec{c} - \vec{d} \cdot \vec{a} \times \vec{c} &= 0 \\ \Rightarrow \vec{c} \cdot \vec{b} \times \vec{a} - \vec{a} \cdot \vec{c} \times \vec{d} &= 0 \quad(2)\\ \Rightarrow -\vec{c} \cdot \vec{a} \times \vec{b} - \vec{a} \cdot \vec{c} \times \vec{d} &= 0 \quad(3)\\ \Rightarrow \vec{a} \cdot \vec{c} \times \vec{d} + \vec{c} \cdot \vec{a} \times \vec{b} &= 0 \quad \text{QED} \end{aligned}$

Notes:

To obtain $(2)$ the cyclic symmetry of the triple scalar product has been used.
To obtain $(3)$ the anti-symmetry of the vector product has been used.

NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

2008 Paper 2 Question 10