Question

The angle between the lines

\[\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{2} \text{ and }, \frac{x - 1}{- \sqrt{3} - 1} = \frac{y - 1}{\sqrt{3} - 1} = \frac{z - 1}{4}\] is 

Options

• (a) \[\cos^{- 1} \left( \frac{1}{65} \right)\]

• (b) \[\frac{\pi}{6}\]

• (c) \[\frac{\pi}{3}\]

• (d) \[\frac{\pi}{4}\]

Answer 1

(c) \[\frac{\pi}{3}\] 

 We have ,

  \[\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{2} \]
\[\frac{x - 1}{- \sqrt{3} - 1} = \frac{y - 1}{\sqrt{3} - 1} = \frac{z - 1}{4}\]

The direction ratios of the given lines are proportional to 1, 1, 2 and   \[- \sqrt{3} - 1, \sqrt{3} - 1, 4\]  The given lines are parallel to vectors  \[\vec{b_1} = \hat{i} + \hat{j} + 2 \hat{k}  \text{ and }  \vec{b_2} = \left( - \sqrt{3} - 1 \right) \hat{i} + \left( \sqrt{3} - 1 \right) \hat{j} + 4 \hat{k}\] Let θ  be the angle between the given lines. 

Now, 

\[\cos \theta = \frac{\vec{b_1} . \vec{b_2}}{\left| \vec{b_1} \right| \left| \vec{b_2} \right|}\]

\[ = \frac{\left( \hat{i} + \hat{j} + 2 \hat{k} \right) . \left\{ \left( - \sqrt{3} - 1 \right) \hat{i}+ \left( \sqrt{3} - 1 \right) \hat{j} + 4 \hat{k}  \right\}}{\sqrt{1^2 + 1^2 + 1^2} \sqrt{\left( - \sqrt{3} - 1 \right)^2 + \left( \sqrt{3} - 1 \right)^2 + 4^2}}\]

\[ = \frac{- \sqrt{3} - 1 + \sqrt{3} - 1 + 8}{\sqrt{3} \sqrt{24}}\]

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

\[ = \frac{1}{\sqrt{2}}\]

\[\]

\[ \Rightarrow \theta = \frac{\pi}{3}\]