Q62 / 180 · KCET 2021 · 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, direction cosines of line $1:\left(l_{1}, m_{1}, n_{1}\right)$
$=\left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}\right)$
direction cosines of line $2 \Rightarrow\left(l_{2}, m_{2}, n_{2}\right)$
$=\left(\frac{\sqrt{3}}{4}, \frac{1}{4},-\frac{\sqrt{3}}{2}\right)$
We know that, angle between two lines is given by $\cos \theta=\left|l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|$
$\begin{aligned}
\cos \theta &=\left|\frac{\sqrt{3}}{4} \times \frac{\sqrt{3}}{4}+\frac{1}{4} \times \frac{1}{4}-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right| \\
&=\left|\frac{3}{16}+\frac{1}{16}-\frac{3}{4}\right|=\left|\frac{3+1-12}{16}\right| \\
&=\left|-\frac{8}{16}\right|=\frac{1}{2} \\
\therefore \quad \theta &=\frac{\pi}{3}
\end{aligned}$
Hence, the correct answer is option: 'C'.