Question

The angle between the lines, whose direction cosines are proportional to 4,√3-1,-√3-1 and 4,-√3-1,√3-1, is

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{2}\)
• E. π

Answer 1

The Correct Option is C

We are given two lines with direction cosines proportional to:

Line 1: \( \langle 4, \sqrt{3} - 1, -\sqrt{3} - 1 \rangle \)

Line 2: \( \langle 4, -\sqrt{3} - 1, \sqrt{3} - 1 \rangle \)

To find the angle \( \theta \) between these lines, we use the dot product formula for direction ratios:

\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

First, compute the dot product of the direction ratios:

\[ 4 \cdot 4 + (\sqrt{3} - 1)(-\sqrt{3} - 1) + (-\sqrt{3} - 1)(\sqrt{3} - 1) \]

\[ = 16 + \left[ -(\sqrt{3})^2 - \sqrt{3} + \sqrt{3} + 1 \right] + \left[ -(\sqrt{3})^2 + \sqrt{3} - \sqrt{3} + 1 \right] \]

\[ = 16 + \left[ -3 + 1 \right] + \left[ -3 + 1 \right] \]

\[ = 16 - 2 - 2 = 12 \]

Next, compute the magnitudes of the direction ratios:

For Line 1:

\[ \sqrt{4^2 + (\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2} \]

\[ = \sqrt{16 + (3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1)} \]

\[ = \sqrt{16 + 4 - 2\sqrt{3} + 4 + 2\sqrt{3}} \]

\[ = \sqrt{24} = 2\sqrt{6} \]

For Line 2:

\[ \sqrt{4^2 + (-\sqrt{3} - 1)^2 + (\sqrt{3} - 1)^2} \]

\[ = \sqrt{16 + (3 + 2\sqrt{3} + 1) + (3 - 2\sqrt{3} + 1)} \]

\[ = \sqrt{16 + 4 + 2\sqrt{3} + 4 - 2\sqrt{3}} \]

\[ = \sqrt{24} = 2\sqrt{6} \]

Now, substitute into the dot product formula:

\[ \cos \theta = \frac{12}{(2\sqrt{6})(2\sqrt{6})} = \frac{12}{24} = \frac{1}{2} \]

Thus, the angle \( \theta \) is:

\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]