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Question
Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.
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Solution
\[\text { Let } \vec{a} \text{ be a vector parallel to the vector with direction ratios } 2, 3, 6 . \]
\[ \Rightarrow \vec{a} = 2 \hat{i} + 3 \hat{j}+ 6 \hat{k} . \]
\[\text{ Let } \vec{b} \text { be a vector parallel to the vector with direction ratios }1, 2, 2 . \]
\[ \Rightarrow \vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k} \]
\[\text { Let } \theta \text{ be the angle between the the given vectors }. \]
\[\text{ Now, }\]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} \]
\[ = \frac{\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right) . \left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right|\left| \hat{i} + 2 \hat{j} + 2 \hat{k} \right|}\]
\[ = \frac{2 + 6 + 12}{\sqrt{4 + 9 + 36} \sqrt{1 + 4 + 4}}\]
\[ = \frac{20}{21} \]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{20}{21} \right)\]
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