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प्रश्न
The angle between the straight lines \[\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} and \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}\] is
पर्याय
a) 45°
(b) 30°
(c) 60°
(d) 90°
उत्तर
(d) 90°
We have ,
\[\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} \]
\[\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}\]
The direction ratios of the given lines are proportional to 2, 5, 4 and 1, 2, -3.
The given lines are parallel to the vectors \[\overrightarrow{b_1} = 2 \hat{i} + 5 \hat{j} + 4 \hat{k} \text{ and } \overrightarrow{b_2} = \hat{i} + 2 \hat{j} - 3 \hat{k} \]
Let θ be the angle between the given lines.
Now,
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( 2 \hat{i} + 5 \hat{j} + 4 \hat{k} \right) . \left( \hat{i} + 2 \hat{j} - 3 \hat{k} \right)}{\sqrt{2^2 + 5^2 + 4^2} \sqrt{1^2 + 2^2 + \left( - 3 \right)^2}}\]
\[ = \frac{2 + 10 - 12}{\sqrt{45} \sqrt{14}}\]
\[ = 0\]
\[ \Rightarrow \theta = 90°\]
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