The direction cosines satisfy the equations:
\[
3l + m + 5n = 0 \quad \text{and} \quad 6m - 2n + 5l = 0
\]
We need to determine the angle \( \theta \) between the two lines. This is achieved by first finding the direction ratios. The given equations are:
3l + m + 5n = 0 ...(i)
and 6mn − 2nl + 5lm = 0 ...(ii)
From (i), rearrange to get m = − 3l − 5n.
Substitute this expression for m into (ii):
6(−3l − 5n)n − 2nl + 5l(−3l − 5n) = 0
This simplifies to (n + l)(2n + l) = 0, which implies either l = −n or l = −2n.
Case 1: If l = −n, substitute into (i) to find m = − 2n.
Case 2: If l = − 2n, substitute into (i) to find m = n.
The direction ratios for the two lines are therefore (−n, −2n, n) and (−2n, n, n). Normalizing these, we obtain the direction ratios 1, 2, −1 and −2, 1, 1.
The direction cosines are related by the formula:
\[
\cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \cdot \sqrt{l_2^2 + m_2^2 + n_2^2}}
\]
Upon solving and simplifying, the angle between the lines is:
\[
\cos \theta = \cos^{-1} \left( -\frac{1}{6} \right)
\]