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Find the Angle Between the Pairs of Lines with Direction Ratios Proportional to 2, 2, 1 and 4, 1, 8 .

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Question

Find the angle between the pairs of lines with direction ratios proportional to  2, 2, 1 and 4, 1, 8 .

 

Sum
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Solution

2, 2, 1 and 4, 1, 8 

\[\text{ Let } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2} \text{ be vectors parallel to the given two lines}  . \]

\[\text{ Then, the angle between the lines is same as the angle between } \overrightarrow{m_1} \text{    and    } \overrightarrow{m_{2 .}} \]

\[\text{ Now }, \]

\[ \overrightarrow{m_1} =\text{  Vector parallel to the line having direction ratios proportional to 2, 2, 1 }\]

\[ \overrightarrow{m_2} = \text{ Vector parallel to the line having direction ratios proportional to 4, 1, 8} \]

\[ \therefore \overrightarrow{m_1} = 2 \hat{i} - 2 \hat{j} + \hat{k} \]

\[ \overrightarrow{m_2} = 4 \hat{i}  + \hat{j}  + 8 \hat{k} \]

\[\text{ Let }  \theta \text{ be the angle between the lines .} \]

\[Now, \]

\[\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}\]

\[ = \frac{\left( 2 \hat{i} + 2 \hat{j} + \hat{k} \right) . \left( 4 \hat{i}+ \hat{j} + 8 \hat{k} \right)}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}}\]

\[ = \frac{8 + 2 + 8}{3 \times 9}\]

\[ = \frac{2}{3}\]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{2}{3} \right)\]

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Chapter 27: Straight Line in Space - Exercise 28.2 [Page 16]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Exercise 28.2 | Q 10.2 | Page 16

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