Question

Prove that: \[\left( \vec{a} - \vec{b} \right) \cdot \left\{ \left( \vec{b} - \vec{c} \right) \times \left( \vec{c} - \vec{a} \right) \right\} = 0\]

Answer 1

We have 

\[( \vec{a} - \vec{b} ) . \left\{ ( \vec{b} \right. - \vec{c} ) \times ( \vec{c} - \vec{a} \left. ) \right\} \]

\[ = ( \vec{a} - \vec{b} ) . ( \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a} ) (\text {By distributive law }) \]

\[ = ( \vec{a} - \vec{b} ) . ( \bar{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a} ) ( \because \vec{c} \times \vec{c} = 0)\]

\[ = ( \vec{a} - \vec{b} ) . ( \bar{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} ) \]

\[ = \vec{a} . ( \vec{b} \times \vec{c} ) + \vec{a} . ( a^\rightharpoonup \times \vec{b} ) + \vec{a} . ( \vec{c} \times \vec{a} ) - \vec{b} . ( \vec{b} \times \vec{c} ) - \vec{b} . ( \vec{a} \times \vec{b} ) - \vec{b} . ( \vec{c} \times \vec{a} ) (\text {By distributive law })\]

\[ = \begin{bmatrix}\vec{a} & \vec{b} & \vec{c}\end{bmatrix} + \begin{bmatrix}\vec{a} & \vec{a} & \vec{b}\end{bmatrix} + \begin{bmatrix}\vec{a} & \vec{c} & \vec{a}\end{bmatrix} - \begin{bmatrix}\vec{b} & \vec{b} & \vec{c}\end{bmatrix} - \left[ \vec{b} \vec{a} \vec{b} \right] - \begin{bmatrix}\vec{b} & \vec{c} & \vec{a}\end{bmatrix}\]

\[ = \begin{bmatrix}\vec{a} & \vec{b} & \vec{c}\end{bmatrix} + 0 + 0 - 0 - 0 - \begin{bmatrix}\vec{a} & \vec{b} & \vec{c}\end{bmatrix} \left( \because \begin{bmatrix}\vec{a} & \vec{b} & \vec{c}\end{bmatrix} = \begin{bmatrix}\vec{b} & \vec{c} & \vec{a}\end{bmatrix} \right) \]

\[ = 0\]