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प्रश्न
The angle between the lines
पर्याय
(a) \[\cos^{- 1} \left( \frac{1}{65} \right)\]
(b) \[\frac{\pi}{6}\]
(c) \[\frac{\pi}{3}\]
(d) \[\frac{\pi}{4}\]
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उत्तर
(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,
\[\cos \theta = \frac{\vec{b_1} . \vec{b_2}}{\left| \vec{b_1} \right| \left| \vec{b_2} \right|}\]
\[ = \frac{\left( \hat{i} + \hat{j} + 2 \hat{k} \right) . \left\{ \left( - \sqrt{3} - 1 \right) \hat{i}+ \left( \sqrt{3} - 1 \right) \hat{j} + 4 \hat{k} \right\}}{\sqrt{1^2 + 1^2 + 1^2} \sqrt{\left( - \sqrt{3} - 1 \right)^2 + \left( \sqrt{3} - 1 \right)^2 + 4^2}}\]
\[ = \frac{- \sqrt{3} - 1 + \sqrt{3} - 1 + 8}{\sqrt{3} \sqrt{24}}\]
\[ = \frac{6}{6\sqrt{2}}\]
\[ = \frac{1}{\sqrt{2}}\]
\[\]
\[ \Rightarrow \theta = \frac{\pi}{3}\]
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