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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.

1 Answer

  • 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.
answered Jun 8, 2013 by thanvigandhi_1

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