If a and b are perpendicular- then a-a-a-a-b is equal to -

Explanation for the correct option:Finding the value of the expression:a×(a×(a×(a×b)))=a×(a×((a.b)a |a|^2b)) [∵x×(x×y)=((x.y)x |x|^2y)] ...

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

If a and b ar...

Question

If $\vec{a}$ and $\vec{b}$ are perpendicular, then $\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b})))$ is equal to :

$(\frac{1}{2}) {|\vec{a}|}^{4} \vec{b}$

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$\vec{a} \times \vec{b}$

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

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$\vec{0}$

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Solution

The correct option is C
${|\vec{a}|}^{4} \vec{b}$

Explanation for the correct option:
Finding the value of the expression:
$\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b}))) = \vec{a} \times (\vec{a} \times ((\vec{a} . \vec{b}) \vec{a} - {|\vec{a}|}^{2} \vec{b}))$ $[∵ \vec{x} \times (\vec{x} \times \vec{y}) = ((\vec{x} . \vec{y}) \vec{x} - {|\vec{x}|}^{2} \vec{y})]$
$= \vec{a} \times (\vec{a} \times (0 - {|\vec{a}|}^{2} \vec{b}))$ …Since dot product of perpendicular vectors is $0$ .
$= \vec{a} \times (- {|\vec{a}|}^{2} (\vec{a} \times \vec{b}))$
$= - {|\vec{a}|}^{2} (\vec{a} \times (\vec{a} \times \vec{b}))$
$= - {|\vec{a}|}^{2} (a (a . b) - {|\vec{a}|}^{2} \vec{b})$
$= - {|\vec{a}|}^{2} (0 - {|\vec{a}|}^{2} \vec{b})$ …Since dot products of perpendicular vectors is $0$ .
$= {|\vec{a}|}^{4} \vec{b}$ .

Hence the correct option is C.

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If a→ and b→ are perpendicular, then a→×a→×a→×a→×b→ is equal to :

If $\vec{a}$ and $\vec{b}$ are perpendicular, then $\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b})))$ is equal to :