If $\vec{a}$, $\vec{b}$, $\vec{c}$ are three non zero vectors such that $\vec{a}\times\vec{b}$ = $\vec{c}$ and $\vec{b}\times\vec{c}$ = $\vec{a}$, prove that $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually at right angles and |$\vec{b}$| = 1 and |$\vec{c}$| = |$\vec{a}$|

Solution :

$\vec{a}\times\vec{b}$ = $\vec{c}$ and $\vec{b}\times\vec{c}$ = $\vec{a}$

$\implies$  $\vec{c}\perp\vec{a}$ , $\vec{c}\perp\vec{b}$ and $\vec{a}\perp\vec{b}$, $\vec{a}\perp\vec{c}$

$\implies$  $\vec{a}\perp\vec{b}$, $\vec{b}\perp\vec{c}$ and $\vec{c}\perp\vec{a}$

$\implies$  $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular vectors.

Again, $\vec{a}\times\vec{b}$ = $\vec{c}$ and $\vec{b}\times\vec{c}$ = $\vec{a}$

$\implies$ |$\vec{a}\times\vec{b}$| = |$\vec{c}$| and |$\vec{b}\times\vec{c}$| = |$\vec{a}$|

$\implies$  $|\vec{a}||\vec{b}|sin{\pi\over 2}$ = |$\vec{c}$| and $|\vec{b}||\vec{c}|sin{\pi\over 2}$ = |$\vec{a}$|  ($\because$ $\vec{a}\perp\vec{b}$ and $\vec{b}\perp\vec{c}$)

$\implies$  $|\vec{a}||\vec{b}|$ = |$\vec{c}$| and $|\vec{b}||\vec{c}|$ = |$\vec{a}$|

$\implies$  ${|\vec{b}|}^2$ |$\vec{c}$| = |$\vec{c}$|

$\implies$  ${|\vec{b}|}^2$ = 1

$\implies$  $|\vec{b}|$ = 1

putting in $|\vec{a}||\vec{b}|$ = |$\vec{c}$|

$\implies$  $|\vec{a}|$ = |$\vec{c}$|

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