# 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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