The direction cosines of two lines are $\left\langle\frac{\sqrt{3}}{2}, \frac{1}{4},…
The direction cosines of two lines are $\left\langle\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right\rangle$ and $\left\langle\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right\rangle$. Then the angle between the lines is equal to
$30^{\circ}$
$60^{\circ}$
$45^{\circ}$
$90^{\circ}$
Solution
Given,
Dr's are $\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ and $\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)$
$\therefore$ Angle between lines is
$
\begin{aligned}
\cos \theta & =\left|l_1 l_2+m_1 m_2+n_1 n_2\right| \\
& \left|\frac{\sqrt{3}}{2}\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4}\left(\frac{1}{4}\right)+\frac{\sqrt{3}}{4}\left(\frac{\sqrt{3}}{4}\right)\right| \\
\cos \theta & \left.=\frac{1}{2} \Rightarrow \frac{3}{16}|=| \frac{-3}{4}+\frac{1}{4} \right\rvert\,
\end{aligned}
$
Hence, option (2) is correct