2008 Paper 2 Question 10

nstmathsupervisor

\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.

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