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The Angle Between the Straight Lines X + 1 2 = Y − 2 5 = Z + 3 4 a N D X − 1 1 = Y + 2 2 = Z − 3 − 3 (A) 45° (B) 30° (C) 60° (D) 90°

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Question

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

Options

  • a) 45°

  • (b) 30°

  • (c) 60°

  • (d) 90°

MCQ
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Solution

(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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Chapter 27: Straight Line in Space - MCQ [Page 42]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
MCQ | Q 1 | Page 42

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