Q41 / 120 · VITEEE 2008 · Mathematics
If $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ be three unit vectors such that $\vec{a} \times(\vec{b} \times \vec{c})=\frac{1}{2} \vec{b}, \vec{b}$ and $\vec{c}$ being non-parallel. If $\theta_{1}$ is the angle between $\vec{a}$ and $\vec{b}$ and $\theta_{2}$ is the angle between $\vec{a}$ and $\vec{c}$, then
- A$\theta_{1}=\frac{\pi}{6}, \theta_{2}=\frac{\pi}{3}$
- B$\theta_{1}=\frac{\pi}{3}, \theta_{2}=\frac{\pi}{6}$
- Correct answerC$\theta_{1}=\frac{\pi}{2}, \theta_{2}=\frac{\pi}{3}$
- D$\theta_{1}=\frac{\pi}{3}, \theta_{2}=\frac{\pi}{2}$
Solution
$\begin{array}{l}
\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{1}{2} \overrightarrow{\mathrm{b}} \\
(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}=\frac{1}{2} \overrightarrow{\mathrm{b}} \quad \text { [Vector triple product] } \\
\left(|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{c}}| \cos \theta_{2}\right) \overrightarrow{\mathrm{b}}-\left(|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \cos \theta_{1}\right) \overrightarrow{\mathrm{c}}=\frac{1}{2} \overrightarrow{\mathrm{b}} \\
\quad[\because|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|=1]
\end{array}$
Equating the coefficients of $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ on both sides, we get
$\begin{array}{l}
\cos \theta_{2}=\frac{1}{2} \text { and }-\cos \theta_{1}=0 \\
\Rightarrow \cos \theta_{2}=\cos \frac{\pi}{3} \text { and } \cos \theta_{1}=\frac{\pi}{2} \\
\Rightarrow \theta_{2}=\frac{\pi}{3} \text { and } \theta_{1}=\frac{\pi}{2}
\end{array}$
Hence, the correct answer is option: 'C'.