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Question: Answered & Verified by Expert
If \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) are three unit vectors such that \(\overrightarrow{\mathrm{b}}\) is not parallel to \(\overrightarrow{\mathrm{c}}\) and \(\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{1}{2} \overrightarrow{\mathrm{b}}\), then the angle between \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{c}}\) is
MathematicsVector AlgebraVITEEEVITEEE 2022
Options:
  • A \(\frac{\pi}{3}\)
  • B \(\frac{\pi}{2}\)
  • C \(\frac{\pi}{6}\)
  • D \(\frac{\pi}{4}\)
Solution:
1782 Upvotes Verified Answer
The correct answer is: \(\frac{\pi}{3}\)
It is given that \(\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{1}{2} \overrightarrow{\mathrm{b}}\)
Note: \(\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}\)
\(\begin{aligned}
& \Rightarrow \quad \frac{1}{2} \overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}} \\
& \Rightarrow \quad\left(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}-\frac{1}{2}\right) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}=0 \quad ...(i)
\end{aligned}\)
Since \(\bar{b}\) and \(\bar{c}\) are non-parallel, therefore for the existence of relation (i), the coeff. of \(\bar{b}\) and \(\overline{\mathrm{c}}\) should vanish separately. Therefore, we get \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}-\frac{1}{2}=0\) i.e., \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\frac{1}{2}\) and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0\) Since \(\vec{a}, \vec{b}, \vec{c}\) are the unit vectors therefore
\(\overline{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \cdot\left(\because \cos ^{-1}\left(\frac{1}{2}\right)=\theta\right)\)

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