Question:
Let \(\vec a,\ \vec b\) and \(\vec c\) be three unit vectors such that \(\vec a\times(\vec b\times\vec c)=\frac{\sqrt3}{2}(\vec b+\vec c)\). If \(\vec b\) is not parallel to \(\vec c\), then the angle between \(\vec a\) and \(\vec b\) is:
Let \(\vec a,\ \vec b\) and \(\vec c\) be three unit vectors such that \(\vec a\times(\vec b\times\vec c)=\frac{\sqrt3}{2}(\vec b+\vec c)\). If \(\vec b\) is not parallel to \(\vec c\), then the angle between \(\vec a\) and \(\vec b\) is:
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Expand the vector triple product and equate coefficients of the independent vectors \(\vec b\) and \(\vec c\).
Updated On: Nov 11, 2026
- \(\frac{2\pi}{3}\)
- \(\frac{5\pi}{6}\)
- \(\frac{3\pi}{4}\)
- \(\frac{\pi}{2}\)
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The Correct Option is B
Solution and Explanation
\(\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c\).
So \((\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c=\frac{\sqrt3}{2}\vec b+\frac{\sqrt3}{2}\vec c\).
Since \(\vec b,\vec c\) are non-parallel (linearly independent), compare coefficients: \(\vec a\cdot\vec c=\frac{\sqrt3}{2}\) and \(\vec a\cdot\vec b=-\frac{\sqrt3}{2}\).
With unit vectors, \(\cos\theta=-\frac{\sqrt3}{2}\Rightarrow\theta=\frac{5\pi}{6}\).
Answer: (2) \(\frac{5\pi}{6}\)
So \((\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c=\frac{\sqrt3}{2}\vec b+\frac{\sqrt3}{2}\vec c\).
Since \(\vec b,\vec c\) are non-parallel (linearly independent), compare coefficients: \(\vec a\cdot\vec c=\frac{\sqrt3}{2}\) and \(\vec a\cdot\vec b=-\frac{\sqrt3}{2}\).
With unit vectors, \(\cos\theta=-\frac{\sqrt3}{2}\Rightarrow\theta=\frac{5\pi}{6}\).
Answer: (2) \(\frac{5\pi}{6}\)
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