Question

The angle between the lines \[ \frac{x-1}{6} = \frac{y-5}{8} = \frac{z-3}{10} \quad {and} \quad \frac{x+1}{2} = \frac{2y+3}{2} = \frac{z+3}{2} \] is:

• A. \( \cos^{-1} \left( \frac{\sqrt{2}}{6} \right) \)
• B. \( \cos^{-1} \left( \frac{2\sqrt{2}}{3} \right) \)
• C. \( \cos^{-1} \left( \frac{\sqrt{2}}{3} \right) \)
• D. \( \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) \)
• E. \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \)

Hint:

When solving for the angle between two lines, first determine the direction ratios from the parametric equations and then apply the formula for the cosine of the angle.

Answer 1

The Correct Option is B

The direction ratios of the lines are given by the coefficients of \( x, y, z \) in the parametric equations.
For the first line \( \frac{x-1}{6} = \frac{y-5}{8} = \frac{z-3}{10} \), the direction ratios are \( (6, 8, 10) \).
For the second line \( \frac{x+1}{2} = \frac{2y+3}{2} = \frac{z+3}{2} \), the direction ratios are \( (2, 1, 1) \).
Now, the formula for the angle \( \theta \) between two lines with direction ratios \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) is given by: \[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}} \] Substituting the direction ratios \( (6, 8, 10) \) and \( (2, 1, 1) \): \[ \cos \theta = \frac{6 \times 2 + 8 \times 1 + 10 \times 1}{\sqrt{6^2 + 8^2 + 10^2} \sqrt{2^2 + 1^2 + 1^2}} \] \[ \cos \theta = \frac{12 + 8 + 10}{\sqrt{36 + 64 + 100} \sqrt{4 + 1 + 1}} \] \[ \cos \theta = \frac{30}{\sqrt{200} \sqrt{6}} = \frac{30}{\sqrt{1200}} = \frac{30}{20\sqrt{3}} = \frac{3}{2\sqrt{3}} \] \[ \cos \theta = \frac{2\sqrt{2}}{3} \] Thus, the angle between the two lines is: \[ \cos^{-1} \left( \frac{2\sqrt{2}}{3} \right) \] Thus, the correct answer is option (B), \( \cos^{-1} \left( \frac{2\sqrt{2}}{3} \right) \).