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Write the Angle Between the Lines X − 5 7 = Y + 2 − 5 = Z − 2 1 a N D X − 1 1 = Y 2 = Z − 1 3 .

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Question

Write the angle between the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z - 2}{1} \text{ and } \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{3} .\]

Short/Brief Note
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Solution

We have , 

\[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z - 2}{1} \]

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

The given lines are parallel to the vectors

\[\overrightarrow{b_1} = 7 \hat{i} - 5 \hat{j} + \hat{k}  \text{ and }  \overrightarrow{b_2} = \hat{i}  + 2 \hat{j}  + 3 \hat{k} \]

Let , 

\[\theta\]  be the angle between the given lines. 

Now ,

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

\[ = \frac{\left( 7 \hat{i}  - 5 \hat{j} + \hat{k}  \right) . \left( \hat{i}  + 2 \hat{j}  + 3 \hat{k}  \right)}{\sqrt{7^2 + \left( - 5 \right)^2 + 1^2} \sqrt{1^2 + 2^2 + 3^2}}\]

\[ = \frac{7 - 10 + 3}{\sqrt{49 + 25 + 1} \sqrt{1 + 4 + 9}}\]

\[ = 0\]

\[ \Rightarrow \theta = \frac{\pi}{2}\]

 

  

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Chapter 27: Straight Line in Space - Very Short Answers [Page 41]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Very Short Answers | Q 9 | Page 41

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