Question

Show that

$[\overline{a}+\overline{b},\overline{b}+\overline{c},\overline{c}+\overline{a}]=2\left[\overline{a}\overline{b}\overline{c}\right]$

Answer 1

$[\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]$

$=\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right|[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$

Let us find the determinant of $\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right|=1(1-0)-1(0-1)+0=1+1=2$

Now, $\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right|[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=2[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$

Hence proved.