Question

If \( \theta \) is the angle between the lines whose direction cosines are given by \( 6mn - 2nl + 5lm = 0 \) and \( 3l + m + 5n = 0 \), then \( \sin \theta = \)}

• A. \( \frac{\sqrt{35}}{6} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{\sqrt{37}}{6} \)
• D. \( \frac{5}{6} \)

Hint:

DRs calculation: substitute one linear equation into the quadratic one to find the ratio of variables.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the angle between two lines whose direction cosines \( (l, m, n) \) satisfy two given equations.
Step 2: Key Formula or Approach:
Substitute \( m = -(3l + 5n) \) from the linear equation into the quadratic equation to find ratios of \( l, m, n \). These ratios will give the direction ratios of the two lines. Then use \( \cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \).
Step 3: Detailed Explanation:
From \( 3l + m + 5n = 0 \), we get \( m = -3l - 5n \).
Substitute into \( 6mn - 2nl + 5lm = 0 \):
\[ 6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0 \] \[ -18nl - 30n^2 - 2nl - 15l^2 - 25ln = 0 \] \[ -15l^2 - 45ln - 30n^2 = 0 \implies l^2 + 3ln + 2n^2 = 0 \] \[ (l + n)(l + 2n) = 0 \implies l = -n \text{ or } l = -2n \] Case 1: If \( l = -n \), then \( m = -3(-n) - 5n = -2n \). DRs: \( (-n, -2n, n) \to (1, 2, -1) \).
Case 2: If \( l = -2n \), then \( m = -3(-2n) - 5n = n \). DRs: \( (-2n, n, n) \to (-2, 1, 1) \).
Calculate \( \cos \theta \):
\[ \cos \theta = \frac{|1(-2) + 2(1) + (-1)(1)|}{\sqrt{1+4+1}\sqrt{4+1+1}} = \frac{|-2 + 2 - 1|}{6} = \frac{1}{6} \] Thus, \( \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{1}{36}} = \frac{\sqrt{35}}{6} \).
Step 4: Final Answer:
The value is \( \frac{\sqrt{35}}{6} \).