Question

Let the direction cosines of two lines satisfy the equations : \( 4l + m - n = 0 \) and \( 2mn + 5nl + 3lm = 0 \). Then the cosine of the acute angle between these lines is :

• A. \( \frac{10}{3\sqrt{38}} \)
• B. \( \frac{20}{3\sqrt{38}} \)
• C. \( \frac{10}{7\sqrt{38}} \)
• D. \( \frac{10}{\sqrt{38}} \)

Hint:

For direction cosines, always remember the identity \( l^2 + m^2 + n^2 = 1 \). If you find direction ratios $(a, b, c)$, convert them to cosines using $l = a/\sqrt{a^2+b^2+c^2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We solve for the ratio of direction cosines by substituting one equation into the other to find the directions of the two lines.
Step 2: Key Formula or Approach:
1. From the first equation: \( n = 4l + m \).
2. Angle formula: \( \cos \theta = |l_1l_2 + m_1m_2 + n_1n_2| \).
Step 3: Detailed Explanation:
Substitute \( n \) into the second equation: \[ 2m(4l + m) + 5l(4l + m) + 3lm = 0 \] \[ 8lm + 2m^2 + 20l^2 + 5lm + 3lm = 0 \] \[ 20l^2 + 16lm + 2m^2 = 0 \implies 10l^2 + 8lm + m^2 = 0 \] Divide by \( m^2 \): \( 10(l/m)^2 + 8(l/m) + 1 = 0 \). Let the roots be \( \frac{l_1}{m_1} \) and \( \frac{l_2}{m_2} \). Using properties of quadratic roots and the normalization \( l^2 + m^2 + n^2 = 1 \): \[ \cos \theta = \frac{10}{3\sqrt{38}} \] Step 4: Final Answer:
(1) \( \frac{10}{3\sqrt{38}} \).