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Question
Find the angle between the pairs of lines with direction ratios proportional to 1, 2, −2 and −2, 2, 1 .
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Solution
1, 2, −2 and −2, 2, 1
\[\text{ Let } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2} \text{ be vectors parallel to the two given lines } . \]
\[ \text{ Then, the angle between the two given lines is same as the angle between } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2 .} \]
\[\text{ Now }, \]
\[ \overrightarrow{m_1} = \text{ Vector parallel to the line having direction ratios proportional to 1, 2, - 2} \]
\[ \overrightarrow{m_2} = \text{ Vector parallel to the line having direction ratios proportional to - 2, 2, 1} \]
\[ \therefore \overrightarrow{m_1} = \hat{i} + 2 \hat{j} - 2 \hat{k} \]
\[ \overrightarrow{m_2} = - 2 \hat{i} + 2 \hat{j} + \hat{k} \]
\[\text{ Let } \theta \text{ be the angle between the lines } . \]
\[Now, \]
\[\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}\]
\[ = \frac{\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) . \left( - 2 \hat{i} + 2 \hat{j} + \hat{k} \right)}{\sqrt{1^2 + 2^2 + \left( - 2 \right)^2} \sqrt{\left( - 2 \right)^2 + 2^2 + 1^2}}\]
\[ = \frac{- 2 + 4 - 2}{3 \times 3}\]
\[ = 0\]
\[ \Rightarrow \theta = \frac{\pi}{2}\]
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