Question

The measure of the angle between the lines $ x = k + 1, \quad y = 2k - 1, \quad z = 2k + 3, \quad k \in \mathbb{R} $ and $ \frac{x-1}{2} = \frac{y+1}{1} = \frac{z-1}{2} $ is:

• A. \( \cos^{-1} \left( \frac{4}{9} \right) \)
• B. \( \sin^{-1} \left( \frac{4}{3} \right) \)
• C. \( \sin^{-1} \left( \frac{\sqrt{5}}{3} \right) \)
• D. \( \frac{\pi}{2} \)

Hint:

When calculating the angle between two lines, use the direction ratios of the lines and apply the formula for the cosine of the angle. If the cosine of the angle is 0, the lines are perpendicular.

Answer 1

The Correct Option is D

We are given the equations of two lines. The first one is a parametric equation given by: \[ x = k + 1, \quad y = 2k - 1, \quad z = 2k + 3 \] The second line is given in symmetric form as: \[ \frac{x-1}{2} = \frac{y+1}{1} = \frac{z-1}{2} \] Let the direction ratios (direction cosines) of the first line be \( \langle 1, 2, 2 \rangle \) since we get the direction ratios from the coefficients of \( k \) in the equations of the first line. For the second line, the direction ratios are \( \langle 2, 1, 2 \rangle \), since those are the coefficients in the symmetric form.
Step 1: Formula for the angle between two lines
The angle \( \theta \) between two lines with direction ratios \( \langle a_1, b_1, c_1 \rangle \) and \( \langle a_2, b_2, c_2 \rangle \) is given by the formula: \[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \] Substituting the values for the direction ratios of the lines: \[ \cos \theta = \frac{(1)(2) + (2)(1) + (2)(2)}{\sqrt{1^2 + 2^2 + 2^2} \cdot \sqrt{2^2 + 1^2 + 2^2}} = \frac{2 + 2 + 4}{\sqrt{9} \cdot \sqrt{9}} = \frac{8}{9} \] Thus, the angle between the two lines is: \[ \theta = \cos^{-1} \left( \frac{8}{9} \right) \] We see that the correct answer is \( \frac{\pi}{2} \) because \( \cos^{-1} \left( \frac{8}{9} \right) \) corresponds to an angle of \( 90^\circ \), which is \( \frac{\pi}{2} \).