Question

Using properties of scalar triple product, prove that $\left[\overline{a}+\overline{b}\overline{b}+\overline{c}\overline{c}+\overline{a}\right]=2\left[\overline{a}\overline{b}\overline{c}\right].$

Answer 1

We have to prove that

$\begin{array}{c}\left[\begin{array}{ccc}\overline{a}+\overline{b}& \overline{b}+\overline{c}& \overline{c}+\overline{a}\end{array}\right]=2\left[\begin{array}{ccc}\overline{a}& \overline{b}& \overline{c}\end{array}\right]\\ \left(\overline{a}+\overline{b}\right).\left(\left(\overline{b}+\overline{c}\right)\left(\overline{c}+\overline{a}\right)\right)\\ \left(\overline{a}+\overline{b}\right).\left(\overline{b}\times \overline{c}+\overline{b}\times \overline{a}+\overline{c}\times \overline{a}\right)\\ \left[\begin{array}{ccc}\overline{a}& \overline{b}& \overline{c}\end{array}\right]+\left[\begin{array}{ccc}\overline{a}& \overline{b}& \overline{c}\end{array}\right]=2\left[\begin{array}{ccc}\overline{a}& \overline{b}& \overline{c}\end{array}\right]\\ \end{array}$

Hence, proved.