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Question: If **u**, **v** and **w** are three non-coplanar vectors, then \((u + v - w).\lbrack(u - v) \times (...

If u, v and w are three non-coplanar vectors, then (u+v−w).[(u−v)×(v−w)](u + v - w).\lbrack(u - v) \times (v - w)\rbrack equals

A

0

B

u.(v×w)u.(v \times w)

C

u.(w×v)u.(w \times v)

D

3u.(v×w)3u.(v \times w)

Answer

u.(v×w)u.(v \times w)

Explanation

Solution

(u+v−w).(u×v−u×w−v×v+v×w)(\mathbf{u} + \mathbf{v} - \mathbf{w}).(\mathbf{u} \times \mathbf{v} - \mathbf{u} \times \mathbf{w} - \mathbf{v} \times \mathbf{v} + \mathbf{v} \times \mathbf{w})

=(u+v−w)(u×v−u×w+v×w)= (\mathbf{u} + \mathbf{v} - \mathbf{w})(\mathbf{u} \times \mathbf{v} - \mathbf{u} \times \mathbf{w} + \mathbf{v} \times \mathbf{w})

=u.(u×v)0−u.(u×w)0+u.(v×w)+v.(u×v)0= \frac{\mathbf{u}.(\mathbf{u} \times \mathbf{v})}{0} - \frac{\mathbf{u}.(\mathbf{u} \times \mathbf{w})}{0} + \mathbf{u}.(\mathbf{v} \times \mathbf{w}) + \frac{\mathbf{v}.(\mathbf{u} \times \mathbf{v})}{0}

−v.(u×w)+v.(v×w)0−w.(u×v)+w.(u×w)0- \mathbf{v}.(\mathbf{u} \times \mathbf{w}) + \frac{\mathbf{v}.(\mathbf{v} \times \mathbf{w})}{0} - \mathbf{w}.(\mathbf{u} \times \mathbf{v}) + \frac{\mathbf{w}.(\mathbf{u} \times \mathbf{w})}{0}

−w.(u×w)0=u.(v×w)−v.(u×w)−w.(u×v)- \frac{\mathbf{w}.(\mathbf{u} \times \mathbf{w})}{0} = \mathbf{u}.(\mathbf{v} \times \mathbf{w}) - \mathbf{v}.(\mathbf{u} \times \mathbf{w}) - \mathbf{w}.(\mathbf{u} \times \mathbf{v})

=[uvw]+[vwu]−[wuv]=u.(v×w).= \lbrack\mathbf{uvw}\rbrack + \lbrack\mathbf{vwu}\rbrack - \lbrack\mathbf{wuv}\rbrack = \mathbf{u}.(\mathbf{v} \times \mathbf{w}).