Q92 / 160 · TS EAMCET 2008 · Mathematics
The angle between the lines whose direction cosines are $\left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}\right)$ and $\left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2}\right)$, is
- A$\pi$
- B$\frac{\pi}{2}$
- Correct answerC$\frac{\pi}{3}$
- D$\frac{\pi}{4}$
Solution
Given that,
$$
l_1=\frac{\sqrt{3}}{4}, m_1=\frac{1}{4} \text { and } n_1=\frac{\sqrt{3}}{2}
$$
and $l_2=\frac{\sqrt{3}}{4}, m_2=\frac{1}{4}$ and $n_2=\frac{-\sqrt{3}}{2}$
$$
\begin{aligned}
\therefore \cos \theta & =\left|l_1 l_2+m_1 m_2+n_1 n_2\right| \\
& =\left|\frac{\sqrt{3}}{4} \times \frac{\sqrt{3}}{4}+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{2} \times\left(\frac{-\sqrt{3}}{2}\right)\right| \\
& =\left|\frac{3}{16}+\frac{1}{16}-\frac{3}{4}\right|=\left|-\frac{2}{4}\right|=\frac{1}{2} \\
\Rightarrow \quad \theta & =\frac{\pi}{3}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.