The angle between the two lines whose direction cosines are given by…

The correct answer is Option (3) → \( \cos^{-1}\left(\displaystyle\frac{19}{3\sqrt{41}}\right) \)

From the linear equation, \( n=-4l-m \). Substituting this in the second equation,

    \( \displaystyle 2m(-4l-m)+24l(-4l-m)+4lm=0 \)

    \( \displaystyle -2m^{2}-28lm-96l^{2}=0 \)

    \( \displaystyle m^{2}+14lm+48l^{2}=0 \)

    \( \displaystyle (m+6l)(m+8l)=0 \)

So \( m=-6l \) or \( m=-8l \), and the corresponding values of n are

    \( \displaystyle m=-6l:\ n=2l \)
    \( \displaystyle m=-8l:\ n=4l \)

The direction ratios of the two lines are \( (1,-6,2) \) and \( (1,-8,4) \), with magnitudes \( \sqrt{41} \) and 9. Therefore

    \( \displaystyle \cos\theta=\frac{1+48+8}{\sqrt{41}\cdot9} \)

    \( \displaystyle =\frac{19}{3\sqrt{41}} \)