Question

The angle between two lines whose direction ratios are proportional to \(1, 1, -2\) and \((\sqrt{3} - 1), (-\sqrt{3} - 1), -4\) is:

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

Hint:

When calculating the angle between two lines in 3D, the dot product and the magnitudes of the direction vectors are crucial. Remember to use the formula \( \cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} \), where \( \vec{d_1} \cdot \vec{d_2} \) is the dot product and \( |\vec{d_1}| \) and \( |\vec{d_2}| \) are the magnitudes of the direction vectors. The angle is then found using the inverse cosine function. Pay close attention to the calculation of magnitudes, as the square roots can sometimes lead to tricky simplifications.

Answer 1

The Correct Option is A

The angle \( \theta \) between two lines whose direction ratios are given by \( \vec{d_1} = (1, 1, -2) \) and \( \vec{d_2} = (\sqrt{3} - 1, -\sqrt{3} - 1, -4) \) can be found using the formula:

\[\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}||\vec{d_2}|}.\]

First, compute the dot product \( \vec{d_1} \cdot \vec{d_2} \):

\[\vec{d_1} \cdot \vec{d_2} = (1)(\sqrt{3} - 1) + (1)(-\sqrt{3} - 1) + (-2)(-4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6.\]

Next, compute the magnitudes of \( \vec{d_1} \) and \( \vec{d_2} \):

\[|\vec{d_1}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6},\]

\[|\vec{d_2}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + (-4)^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{24} = 2\sqrt{6}.\]

Thus, we have:

\[\cos \theta = \frac{6}{\sqrt{6} \times 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2}.\]

Therefore,

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

Thus, the correct answer is:

\(\frac{\pi}{3}\).

Answer 2

The angle \( \theta \) between two lines whose direction ratios are given by \( \vec{d_1} = (1, 1, -2) \) and \( \vec{d_2} = (\sqrt{3} - 1, -\sqrt{3} - 1, -4) \) can be found using the formula:

\[ \cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}||\vec{d_2}|}. \]

Step 1: Compute the dot product \( \vec{d_1} \cdot \vec{d_2} \):

\[ \vec{d_1} \cdot \vec{d_2} = (1)(\sqrt{3} - 1) + (1)(-\sqrt{3} - 1) + (-2)(-4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6. \]

Step 2: Compute the magnitudes of \( \vec{d_1} \) and \( \vec{d_2} \):

\[ |\vec{d_1}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}, \] \[ |\vec{d_2}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + (-4)^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{24} = 2\sqrt{6}. \]

Step 3: Compute \( \cos \theta \):

\[ \cos \theta = \frac{6}{\sqrt{6} \times 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2}. \]

Step 4: Compute the angle \( \theta \):

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

Thus, the correct answer is:

\[ \frac{\pi}{3}. \]