Question

The angle between the lines \[ \frac{x-1}{1} = \frac{y+1}{2} = \frac{z-3}{-1}, \quad \frac{x-1}{2} = \frac{y-3}{3} = \frac{z-1}{4}, \] where \( l, m, n \) are roots of the equation \[ x^2 + x^2 - 4x - 4 = 0, \] is

• A. \( \cos^{-1} \left( \frac{2}{3} \right) \)
• B. \( \cos^{-1} \left( \frac{1}{3} \right) \)
• C. \( \cos^{-1} \left( \frac{3}{4} \right) \)
• D. \( \cos^{-1} \left( \frac{1}{4} \right) \)

Hint:

To find the angle between two lines, use the formula involving the dot product of their direction vectors and their magnitudes.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We first need to find the values of \( l, m, n \) by solving the cubic equation, then calculate the angle between lines with direction ratios \( (l, m, n) \) and \( (m, n, l) \).
Step 2: Detailed Explanation:
1. Solve \( x^3 + x^2 - 4x - 4 = 0 \):
\( x^2(x + 1) - 4(x + 1) = 0 \Rightarrow (x^2 - 4)(x + 1) = 0 \).
Roots are \( 2, -2, -1 \).
Given \( l>m>n \), so \( l = 2, m = -1, n = -2 \).
2. Direction vectors:
\( \vec{b_1} = (2, -1, -2) \)
\( \vec{b_2} = (-1, -2, 2) \)
3. Angle \( \cos \theta = \frac{l \cdot m + m \cdot n + n \cdot l}{l^2 + m^2 + n^2} \):
\( \cos \theta = \frac{(2)(-1) + (-1)(-2) + (-2)(2)}{2^2 + (-1)^2 + (-2)^2} \)
\( \cos \theta = \frac{-2 + 2 - 4}{9} = \frac{-4}{9} \).
\( \theta = \cos^{-1}(-4/9) \).
Step 3: Final Answer:
The angle is \( \cos^{-1} \left(\frac{-4}{9}\right) \).