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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 .
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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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