Question

If the direction cosines of two lines satisfy the equations $ l - 2m + n = 0 $ and $ lm + 10mn - 2nl = 0 $, and $ \theta $ is the angle between the lines, then $ \cos \theta = $

• A. $ \frac{\pi}{6} $
• B. $ \frac{8}{\sqrt{70}} $
• C. $ \frac{\pi}{3} $
• D. $ \frac{20}{3\sqrt{70}} $

Hint:

When solving problems involving direction cosines and angles between lines, use the given equations to express one variable in terms of another, substitute, and simplify systematically. Use the formula for the cosine of the angle between two lines to find the answer.

Answer 1

The Correct Option is B

Step 1: Direction cosines and equations.
The direction cosines of the lines satisfy:
1. \( l - 2m + n = 0 \),
2. \( lm + 10mn - 2nl = 0 \).
From the first equation: $$ l = 2m - n. $$ Substitute \( l = 2m - n \) into the second equation: $$ (2m - n)m + 10mn - 2n(2m - n) = 0. $$ Simplify: $$ 2m^2 - mn + 10mn - 4mn + 2n^2 = 0, $$ $$ 2m^2 + 5mn + 2n^2 = 0. $$ Factorize: $$ (2m + n)(m + 2n) = 0. $$ Thus: $$ 2m + n = 0 \quad \text{or} \quad m + 2n = 0. $$ Step 2: Solve for direction cosines.
Case 1: \( 2m + n = 0 \Rightarrow n = -2m \).
Substitute \( n = -2m \) into \( l = 2m - n \): $$ l = 2m - (-2m) = 4m. $$ Thus, the direction ratios are proportional to \( (4m, m, -2m) \), or simplified as \( (4, 1, -2) \).
Case 2: \( m + 2n = 0 \Rightarrow m = -2n \).
Substitute \( m = -2n \) into \( l = 2m - n \): $$ l = 2(-2n) - n = -4n - n = -5n. $$ Thus, the direction ratios are proportional to \( (-5n, -2n, n) \), or simplified as \( (-5, -2, 1) \).
Step 3: Cosine of the angle between the lines.
The cosine of the angle \( \theta \) between two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) is given by: $$ \cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}. $$ For direction ratios \( (4, 1, -2) \) and \( (-5, -2, 1) \): $$ \cos \theta = \frac{(4)(-5) + (1)(-2) + (-2)(1)}{\sqrt{4^2 + 1^2 + (-2)^2} \cdot \sqrt{(-5)^2 + (-2)^2 + 1^2}}. $$ Simplify the numerator: $$ (4)(-5) + (1)(-2) + (-2)(1) = -20 - 2 - 2 = -24. $$ Simplify the denominators: $$ \sqrt{4^2 + 1^2 + (-2)^2} = \sqrt{16 + 1 + 4} = \sqrt{21}, $$ $$ \sqrt{(-5)^2 + (-2)^2 + 1^2} = \sqrt{25 + 4 + 1} = \sqrt{30}. $$ Thus: $$ \cos \theta = \frac{-24}{\sqrt{21} \cdot \sqrt{30}} = \frac{-24}{\sqrt{630}} = \frac{-24}{3\sqrt{70}} = -\frac{8}{\sqrt{70}}. $$ Since \( \cos \theta \) must be positive (as angles between lines are measured in \( [0, \pi] \)): $$ \cos \theta = \frac{8}{\sqrt{70}}. $$ Step 4: Final Answer.
$$ \boxed{\frac{8}{\sqrt{70}}} $$