Question

The angle between the lines \[ \frac{x-1}{2} = \frac{2y+3}{4} = \frac{z+5}{-2} \quad {and} \quad \frac{x-3}{4} = \frac{y+1}{-4} = \frac{z+3}{-4} \] is equal to

• A. \( \cos^{-1} \left(\frac{1}{8} \right) \)
• B. \( \cos^{-1} \left(\frac{1}{3} \right) \)
• C. \( \cos^{-1} \left(\frac{1}{4} \right) \)
• D. \( \cos^{-1} \left(\frac{1}{12} \right) \)
• E. \( \cos^{-1} \left(\frac{1}{\sqrt{3}} \right) \)

Hint:

To find the angle between two lines in 3D, use the formula \( \cos \theta = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}| |\mathbf{d_2}|} \).

Answer 1

The Correct Option is B

The angle \( \theta \) between two lines given in symmetric form \[ \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}, \quad \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} \] is given by: \[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \] Computing the dot product and magnitudes from the given equations, we obtain: \[ \cos \theta = \frac{1}{3} \] \[ \theta = \cos^{-1} \left(\frac{1}{3}\right) \] Thus, the correct answer is (B).