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If \left [ \vec{a}\times \vec{b}\: \: \vec{b} \times \vec{c}\: \: \vec{c} \times \vec{a} \right ]= \lambda\left [\vec{a}\: \vec{b}\: \vec{c}\right ]^{2}then \lambda is equal to:

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best_answer

As we learnt in 

Scalar Triple Product -

\left [ \vec{a}\;\vec{b}\; \vec{c} \right ]

=\left (\vec{a}\times \vec{b}\right)\cdot \vec{c}= \vec{a}\cdot \left ( \vec{b} \times \vec{c}\right )

=\left (\vec{b}\times \vec{c}\right)\cdot \vec{a}= \vec{b}\cdot \left ( \vec{c} \times \vec{a}\right )

=\left (\vec{c}\times \vec{a}\right)\cdot \vec{b}= \vec{c}\cdot \left ( \vec{a} \times \vec{b}\right )

- wherein

Scalar Triple Product of three vectors \hat{a},\hat{b},\hat{c}.

 

 \left [ \vec{a} \right \times \vec{b} \:\:\:\vec{b} \right \times \vec{c}\:\:\: \vec{c} \right \times \vec{a}]=\lambda \left [ \vec{a} \right\vec{b} \vec{c}]^{2}

\Rightarrow (\vec{a}\times \vec{b}).((\vec{b}\times \vec{c})\times (\vec{c}\times \vec{a}))

\Rightarrow (\vec{a}\times \vec{b}).((\vec{b}\times \vec{c}).\vec{a})\vec{c}-((\vec{b}\times\vec{c}).\vec{c})\vec{a}

\Rightarrow (\vec{a}\times \vec{b}).(\left [ \right \vec{b}\vec{c}\vec{a}]\vec{c}-0)

=\left [ \vec{a} \right\vec{b} \vec{c} ]\times \left [ \vec{b} \right\vec{c} \vec{a} ]

=\left [ \vec{a} \right \vec{b} \vec{c} ]^{2}

Thus \lambda =1

 

 


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