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. $\frac{\pi}{2}$

Answer 1

Answer: (A)

$\frac{\pi}{3}$

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:

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

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:

$\frac{\pi}{3}$.