# Prove that $\mid[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]\mid=a b c$ if and only if $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular.

Toolbox:
• If $\overrightarrow a \perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$ and for nonzero vectors if $\overrightarrow a.\overrightarrow b=0 \Rightarrow \overrightarrow a \perp \overrightarrow b.$
• If $\overrightarrow a \parallel \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=|\overrightarrow a|\: |\overrightarrow b|$ if they are in the same direction and $\overrightarrow a.\overrightarrow b=-|\overrightarrow a||\overrightarrow b|$ if they are in opposite directions.
• $( \overrightarrow a \times \overrightarrow b).\overrightarrow c = (\overrightarrow b \times \overrightarrow c)\overrightarrow a = (\overrightarrow c \times \overrightarrow a).\overrightarrow b = [ \overrightarrow a, \overrightarrow b, \overrightarrow c ]$ Also $[ \overrightarrow a\: \overrightarrow b\: \overrightarrow c] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$ = volume of the parallelopiped formed by three non-coplanar vectors $\overrightarrow a, \overrightarrow b, \overrightarrow c$.
Step 1
Let $\overrightarrow a , \overrightarrow b, \overrightarrow c$ be mutually perpendicular.
Then $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow c = \overrightarrow c.\overrightarrow a=0$
Consider $[ \overrightarrow a\: \overrightarrow b\: \overrightarrow c]=\overrightarrow a.(\overrightarrow b\times\overrightarrow c)$
Since $\overrightarrow a \perp \overrightarrow b \: and \: \overrightarrow a \perp \overrightarrow c, \: \overrightarrow a \parallel \overrightarrow b \times \overrightarrow c$
$\therefore \overrightarrow a(\overrightarrow b \times \overrightarrow c) = \pm |\overrightarrow a||\overrightarrow b \times \overrightarrow c|$
$= \pm |\overrightarrow a||\overrightarrow b||\overrightarrow c| \sin \large\frac{\pi}{2} \: (since\: \overrightarrow b \perp \overrightarrow c)$
$= \pm abc\: \: or \: |[\overrightarrow a\: \overrightarrow b\: \overrightarrow c]| = abc$
Conversely let $|[\overrightarrow a\: \overrightarrow b\: \overrightarrow c ]|=abc$
Step 2
$\Rightarrow [\overrightarrow a\: \overrightarrow b\: \overrightarrow c ] = \pm abc$
$\Rightarrow \overrightarrow a.(\overrightarrow b \times \overrightarrow c) = \pm abc$
$\Rightarrow \overrightarrow a \parallel (\overrightarrow b \times \overrightarrow c) \: or \: \overrightarrow a \perp$ to the plane of $\overrightarrow b\: and \: \overrightarrow c \: and \: \overrightarrow a \perp \overrightarrow b, \overrightarrow a \perp \overrightarrow c$ (i)
Parallely $[\overrightarrow a\: \overrightarrow b\: \overrightarrow c] = (\overrightarrow a \times \overrightarrow b).\overrightarrow c$
$(\overrightarrow a \times \overrightarrow b).\overrightarrow c = \pm abc$
$\Rightarrow (\overrightarrow a \times \overrightarrow b) \parallel \overrightarrow c \: or \: \overrightarrow c \: is \: \perp$ to the plane $\overrightarrow a \: and \: \overrightarrow b, \overrightarrow c \perp \overrightarrow a \: and \: \overrightarrow c \perp \overrightarrow b.$ (ii)
From (i) and (ii) it follows that $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are mutually perpendicular.