Find the angle between the pair of lines $\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4}$ and $\frac{x+1}{1} = \frac{y-4}{-1} = \frac{z-5}{2}$.

$\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$

The correct answer is Option (1) → $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$ ##

The direction ratios of the first line are $3, 5, 4$ and the direction ratios of the second line are $1, 1, 2$. If $\theta$ is the angle between them, then:

$\cos \theta = \left| \frac{3 \cdot 1 + 5 \cdot 1 + 4 \cdot 2}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 1^2 + 2^2}} \right|$

$= \frac{16}{\sqrt{50} \sqrt{6}} = \frac{16}{5\sqrt{2} \cdot \sqrt{6}} = \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}} = \frac{8\sqrt{3}}{15}$

Hence, the required angle is $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$.