Question:

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

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Use \(\vec a\times(\vec a\times\vec v)=(\vec a\cdot\vec v)\vec a-|\vec a|^2\vec v\) repeatedly; with \(\vec a\perp\vec b\) the dot terms vanish.
Updated On: Nov 4, 2026
  • \(\frac{1}{2}|\vec{a}|^4\vec{b}\)
  • \(\vec{0}\)
  • \(\vec{a}\times\vec{b}\)
  • \(|\vec{a}|^4\vec{b}\)
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The Correct Option is D

Solution and Explanation

For \(\vec a\perp\vec b\): \(\vec a\times(\vec a\times\vec b)=(\vec a\cdot\vec b)\vec a-|\vec a|^2\vec b=-|\vec a|^2\vec b\).
Then \(\vec a\times\left(\vec a\times(\vec a\times\vec b)\right)=\vec a\times\left(-|\vec a|^2\vec b\right)=-|\vec a|^2(\vec a\times\vec b)\).
Finally \(\vec a\times\left[-|\vec a|^2(\vec a\times\vec b)\right]=-|\vec a|^2\left(\vec a\times(\vec a\times\vec b)\right)=-|\vec a|^2\left(-|\vec a|^2\vec b\right)=|\vec a|^4\vec b\).
Answer: (4) \(|\vec{a}|^4\vec{b}\)
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