Question

If a line \( L \) makes angles \( \frac{\pi}{3} \) and \( \frac{\pi}{4} \) with the Y-axis and Z-axis respectively, then the angle between \( L \) and another line having direction ratios \( 1,1,1 \) is:

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

Hint:

The angle between two lines with direction cosines \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) is given by: \[ \cos\theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \cdot \sqrt{l_2^2 + m_2^2 + n_2^2}}. \]

Answer 1

The Correct Option is D

Step 1: Find Direction Ratios of Line \( L \)
The direction cosines \( l, m, n \) of line \( L \) are given by: \[ m = \cos\frac{\pi}{3}, \quad n = \cos\frac{\pi}{4}. \] \[ m = \frac{1}{2}, \quad n = \frac{1}{\sqrt{2}}. \] Using the identity: \[ l^2 + m^2 + n^2 = 1, \] \[ l^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1. \] \[ l^2 + \frac{1}{4} + \frac{1}{2} = 1. \] \[ l^2 + \frac{3}{4} = 1. \] \[ l^2 = \frac{1}{4} \Rightarrow l = \frac{1}{2}. \] Thus, the direction ratios of \( L \) are proportional to: \[ \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}} \right). \] Step 2: Find the Angle Between the Two Lines
The direction ratios of the second line are \( (1,1,1) \). The angle \( \theta \) between the lines is given by: \[ \cos\theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \cdot \sqrt{l_2^2 + m_2^2 + n_2^2}}. \] Substituting values: \[ \cos\theta = \frac{\left(\frac{1}{2} \times 1\right) + \left(\frac{1}{2} \times 1\right) + \left(\frac{1}{\sqrt{2}} \times 1\right)}{\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} \cdot \sqrt{1^2 + 1^2 + 1^2}}. \] \[ = \frac{\frac{1}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}}{\sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{2}} \cdot \sqrt{3}}. \] \[ = \frac{1 + \frac{1}{\sqrt{2}}}{\sqrt{1} \cdot \sqrt{3}} = \frac{\sqrt{2} + 1}{\sqrt{6}}. \] Step 3: Conclusion
Thus, the correct answer is: \[ \mathbf{\cos^{-1} \left( \frac{\sqrt{2}+1}{\sqrt{6}} \right)}. \]