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Question
Find the angle between the pairs of lines with direction ratios proportional to 5, −12, 13 and −3, 4, 5
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Solution
5, −12, 13 and −3, 4, 5
\[\text{ Let } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2} \text{ be vectors parallel to the two given lines } . \]
\[\text { Then, the angle between the two given lines is same as the angle between } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2} . \]
\[\text{ Now }, \]
\[ \overrightarrow{m_1} = \text{ Vector parallel to the line having direction ratios proportional to } 5, - 12, 13\]
\[ \overrightarrow{m_2} = \text{ Vector parallel to the line having direction ratios proportional to } - 3, 4, 5\]
\[ \therefore \overrightarrow{m_1} = 5 \hat{i} - 12 \hat{j} + 13 \hat{k} \]
\[ \overrightarrow{m_2} = - 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \]
\[\text{ Let } \theta \text{ be the angle between the lines } \].
\[\text{ Now } , \]
\[\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}\]
\[ = \frac{\left( 5 \hat{i} - 12 \hat{j} + 13 \hat{k} \right) . \left( - 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)}{\sqrt{5^2 + \left( - 12 \right)^2 + {13}^2} \sqrt{\left( - 3 \right)^2 + 4^2 + 5^2}}\]
\[ = \frac{- 15 - 48 + 65}{13\sqrt{2} \times 5\sqrt{2}}\]
\[ = \frac{1}{65}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{65} \right)\]
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