Question

The angle between the lines whose direction cosines are given by the equations \( 3l + m + 5n = 0 \) and \( 6m - 2n + 5l = 0 \) is:

• A. \( \cos^{-1} \left( \frac{1}{6} \right) \)
• B. \( \cos^{-1} \left( -\frac{1}{6} \right) \)
• C. \( \cos^{-1} \left( \frac{2}{3} \right) \)
• D. \( \cos^{-1} \left( -\frac{5}{6} \right) \)

Hint:

To calculate the angle between two lines, first find their direction ratios and then apply the cosine formula.

Answer 1

The Correct Option is B

The given equations for direction cosines are: \[ 3l + m + 5n = 0 \quad {and} \quad 6m - 2n + 5l = 0 \] From the given, we need to find the angle \( \theta \) between the two lines. To do so, first, solve these equations for the direction ratios and use the formula for the cosine of the angle between two lines: 3l + m + 5n = 0 ...(i)
and 6mn − 2nl + 5lm = 0 ...(ii)
From (i), we have m = − 3l − 5n.
Putting m = − 3l − 5n in (ii),
we get 6(−3l − 5n)n − 2nl + 5l(−3l − 5n) = 0
⇒ (n + l)(2n + l) = 0
⇒ either l = −n or l = −2n.
If l = − n, then putting l = −n in (i), we obtain m = − 2n.
If l = − 2n, then putting l = − 2n in (i), we obtain m = n.
Thus, the direction ratios of two lines are −n, − 2n,
n and −2n,n,n i.e., 1,2,−1 and −2,1,1.
Hence, the direction cosines are
\[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \cdot \sqrt{l_2^2 + m_2^2 + n_2^2}} \] After solving the system of equations and simplifying, the angle between the lines is given by: \[ \cos \theta = \cos^{-1} \left( -\frac{1}{6} \right) \]