If a¯ and b¯ are perpendicular. Then a¯ ×a¯ ×a¯×a¯×b¯ is equal to

a¯⋅b¯=0,a¯×(a¯×b¯)=0(a¯⋅b¯)a¯−(a¯×a¯)b¯=0−|a|2b¯=|a|2b¯a¯×(a¯×−|a¯|2b)=−|a¯|2(a¯×(a¯×b¯))=−|a¯|2(−|a¯|2b)=|a¯|4b¯

If a¯ and b¯ are perpendicular. Then a¯ ×a¯ ×a¯×a¯×b¯ is equal to

a¯⋅b¯=0,a¯×(a¯×b¯)=0(a¯⋅b¯)a¯−(a¯×a¯)b¯=0−|a|2b¯=|a|2b¯a¯×(a¯×−|a¯|2b)=−|a¯|2(a¯×(a¯×b¯))=−|a¯|2(−|a¯|2b)=|a¯|4b¯

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If $\bar{a}$ and $\bar{b}$ are perpendicular. Then $\bar{a} \times (\bar{a} \times (\bar{a} \times (\bar{a} \times \bar{b})))$ is equal to

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$\bar{O}$

$\frac{1}{2} {|\bar{a}|}^{4} |\bar{b}|$

$\bar{a} \times \bar{b}$

$|\bar{a}|^{4} \bar{b}$

answer is D.

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Detailed Solution

$\bar{a} \cdot \bar{b} = 0, \bar{a} \times (\bar{a} \times \bar{b}) = 0 (\bar{a} \cdot \bar{b}) \bar{a} - (\bar{a} \times \bar{a}) \bar{b}$

$= 0 - | a |^{2} \bar{b} = | a |^{2} \bar{b} \bar{a} \times (\bar{a} \times - | \bar{a} |^{2} b) = - | \bar{a} |^{2} (\bar{a} \times (\bar{a} \times \bar{b}))$

$= - | \bar{a} |^{2} (- | \bar{a} |^{2} b) = | \bar{a} |^{4} \bar{b}$

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If a¯ and b¯ are perpendicular. Then a¯ ×a¯ ×a¯×a¯×b¯ is equal to

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If $\bar{a}$ and $\bar{b}$ are perpendicular. Then $\bar{a} \times (\bar{a} \times (\bar{a} \times (\bar{a} \times \bar{b})))$ is equal to