Using properties of scalar triple product, prove that \([\bar{a} + \bar{b}\ \ \bar{b} + \bar{c}\ \ \bar{c} + \bar{a}] = 2[\bar{a}\ \bar{b}\ \bar{c}]\)

Answer: Proof by expanding the scalar triple product.

LHS \(= (\bar{a} + \bar{b})\cdot[(\bar{b} + \bar{c}) \times (\bar{c} + \bar{a})]\).

\((\bar{b} + \bar{c}) \times (\bar{c} + \bar{a}) = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{c} + \bar{c} \times \bar{a} = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a}\) (since \(\bar{c} \times \bar{c} = \bar{0}\)).

\(\therefore\) LHS \(= \bar{a}\cdot(\bar{b} \times \bar{c}) + \bar{a}\cdot(\bar{b} \times \bar{a}) + \bar{a}\cdot(\bar{c} \times \bar{a}) + \bar{b}\cdot(\bar{b} \times \bar{c}) + \bar{b}\cdot(\bar{b} \times \bar{a}) + \bar{b}\cdot(\bar{c} \times \bar{a})\)

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

A scalar triple product with two equal vectors is zero, so the 2nd to 5th terms vanish. Also \([\bar{b}\ \bar{c}\ \bar{a}] = [\bar{a}\ \bar{b}\ \bar{c}]\) (cyclic property).

\(\therefore\) LHS \(= [\bar{a}\ \bar{b}\ \bar{c}] + [\bar{a}\ \bar{b}\ \bar{c}] = 2[\bar{a}\ \bar{b}\ \bar{c}] =\) RHS.