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If  \overrightarrow{a} and \overrightarrow{b} are perpendicular, then \overrightarrow{a} \times \left ( \overrightarrow{a} \times \left ( \overrightarrow{a} \times \left ( \overrightarrow{a} \times \overrightarrow{b} \right ) \right ) \right ) is equal to :
 
Option: 1 \overrightarrow{0}
Option: 2 \overrightarrow{a}\times \overrightarrow{b}
Option: 3 \left | \overrightarrow{a} \right |^{4}\overrightarrow{b}  
Option: 4 \frac{1}{2}\left | \overrightarrow{a} \right |^{4}\overrightarrow{b}

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best_answer

Given that

\\\vec{a} \text { and } \vec{b} \text { are perpendicular }\\\text{So, }\vec a\cdot \vec b=0\\\vec{\mathrm{a}} \times(\vec{\mathrm{a}} \times \vec{\mathrm{b}})=(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}) \vec{\mathrm{a}}-(\vec{\mathrm{a}} \cdot \vec{\mathrm{a}}) \vec{\mathrm{b}}=-|\vec{\mathrm{a}}|^{2} \vec{\mathrm{b}}

\\ \text { Now } \vec{\mathrm{a}} \times\left(\vec{\mathrm{a}} \times\left(-|\vec{\mathrm{a}}|^{2} \vec{\mathrm{b}}\right)\right) \\ =\left ( \vec a\cdot \left ( -|\vec a|^2 \vec b\right ) \right )\cdot \vec a-\left ( \vec a\cdot \vec a \right )\cdot\left ( -|\vec a|^2 \vec b\right ) \\ =|\vec a|^2\left (\left ( \vec a\cdot \vec b \right )\cdot \vec a+|\vec a|^2\vec b \right ) \\ =|\vec a|^4\vec b

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himanshu.meshram

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