# For any three vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ prove that [$$\vec{a}$$ + $$\vec{b}$$ $$\vec{b}$$ + $$\vec{c}$$ $$\vec{c}$$ + $$\vec{a}$$] = 2[$$\vec{a}$$ $$\vec{b}$$ $$\vec{c}$$]

## Solution :

We have [$$\vec{a}$$ + $$\vec{b}$$ $$\vec{b}$$ + $$\vec{c}$$ $$\vec{c}$$ + $$\vec{a}$$]

= {($$\vec{a}$$ + $$\vec{b}$$)$$\times$$($$\vec{b}$$ + $$\vec{c}$$)}.($$\vec{c}$$ + $$\vec{a}$$)

= {$$\vec{a}$$$$\times$$$$\vec{b}$$ + $$\vec{a}$$$$\times$$$$\vec{c}$$ + $$\vec{b}$$$$\times$$$$\vec{b}$$ + $$\vec{b}$$$$\times$$$$\vec{c}$$}.($$\vec{c}$$ + $$\vec{a}$$)  {$$\vec{b}$$$$\times$$$$\vec{b}$$ = 0}

= {$$\vec{a}$$$$\times$$$$\vec{b}$$ + $$\vec{a}$$$$\times$$$$\vec{c}$$ + $$\vec{b}$$$$\times$$$$\vec{c}$$}.($$\vec{c}$$ + $$\vec{a}$$)

= ($$\vec{a}\times\vec{b}$$).$$\vec{c}$$ + ($$\vec{a}\times\vec{c}$$).$$\vec{c}$$ + ($$\vec{b}\times\vec{c}$$).$$\vec{c}$$ + ($$\vec{a}\times\vec{b}$$).$$\vec{a}$$ + ($$\vec{a}\times\vec{c}$$).$$\vec{a}$$ + ($$\vec{b}\times\vec{c}$$).$$\vec{a}$$

= [$$\vec{a}$$ $$\vec{b}$$ $$\vec{c}$$] + 0 + 0 + 0 + 0 + [$$\vec{b}$$ $$\vec{c}$$ $$\vec{a}$$]

= [$$\vec{a}$$ $$\vec{b}$$ $$\vec{c}$$] + [$$\vec{a}$$ $$\vec{b}$$ $$\vec{c}$$] = 2[$$\vec{a}$$ $$\vec{b}$$ $$\vec{c}$$]

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