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Question
Find the angle between the vectors whose direction cosines are proportional to 2, 3, −6 and 3, −4, 5.
Solution
\[\text{ Let } \vec{a} \text{ be a vector with direction ratios } 2, 3, - 6 . \]
\[ \Rightarrow \vec{a} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} .\]
\[\ \text { Let } \vec{b} \text { be a vector with direction ratios } 3, - 4, 5 . \]
\[ \Rightarrow \vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \]
\[\text{ Let } \theta \text{ be the angle between the given vectors } . \]
\[\text{ Now, }\]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} \]
\[ = \frac{\left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) . \left( 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right|\left| 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \right|}\]
\[ = \frac{6 - 12 - 30}{\sqrt{4 + 9 + 36} \sqrt{9 + 16 + 25}} \]
\[ = \frac{- 36}{\sqrt{49} \sqrt{50}} \]
\[ = \frac{- 36}{35\sqrt{2}}\]
\[\text{ Rationalising the result, we get }\]
\[\cos \theta = - \frac{18\sqrt{2}}{35} \]
\[ \therefore \theta = \cos^{- 1} \left( - \frac{18\sqrt{2}}{35} \right)\]
\[\ \text { Thus, the angle between the given vectors measures }\cos^{- 1} \left( - \frac{18\sqrt{2}}{35} \right) . \]
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