1.II .5B. 5 B \quad

Vectors and Matrices | Part IA, 2008

(a) Use suffix notation to prove that

a×(b×c)=(ac)b(ab)c\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}

Hence, or otherwise, expand (i) (a×b)(c×d)(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d}), (ii) (a×b)[(b×c)×(c×a)](\mathbf{a} \times \mathbf{b}) \cdot[(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})].

(b) Write down the equation of the line that passes through the point a and is parallel to the unit vector t^\hat{\mathbf{t}}.

The lines L1L_{1} and L2L_{2} in three dimensions pass through a1\mathbf{a}_{1} and a2\mathbf{a}_{2} respectively and are parallel to the unit vectors t^1\hat{\mathbf{t}}_{1} and t^2\hat{\mathbf{t}}_{2} respectively. Show that a necessary condition for L1L_{1} and L2L_{2} to intersect is

(a1a2)(t^1×t^2)=0\left(\mathbf{a}_{1}-\mathbf{a}_{2}\right) \cdot\left(\hat{\mathbf{t}}_{1} \times \hat{\mathbf{t}}_{2}\right)=0

Why is this condition not sufficient?

In the case in which L1L_{1} and L2L_{2} are non-parallel and non-intersecting, find an expression for the shortest distance between them.

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