Question

Find the angle between the vectors whose direction cosines are proportional to 2, 3, −6 and 3, −4, 5.

Answer 1

\[\text{ Let } \vec{a} \text{ be a vector with direction ratios } 2, 3, - 6 . \]

\[ \Rightarrow \vec{a} = 2  \hat{i} + 3 \hat{j} - 6 \hat{k} .\]

\[\ \text { Let } \vec{b} \text { be a vector with direction ratios }  3, - 4, 5 . \]

\[ \Rightarrow \vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \]

\[\text{ Let } \theta \text{  be the angle between the given vectors }  . \]

\[\text{ Now, }\]

\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} \]

\[ = \frac{\left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) . \left( 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right|\left| 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \right|}\]

\[ = \frac{6 - 12 - 30}{\sqrt{4 + 9 + 36} \sqrt{9 + 16 + 25}} \]

\[ = \frac{- 36}{\sqrt{49} \sqrt{50}} \]

\[ = \frac{- 36}{35\sqrt{2}}\]

\[\text{ Rationalising the result, we get }\]

\[\cos \theta = - \frac{18\sqrt{2}}{35} \]

\[ \therefore \theta = \cos^{- 1} \left( - \frac{18\sqrt{2}}{35} \right)\]

\[\ \text { Thus, the angle between the given vectors measures }\cos^{- 1} \left( - \frac{18\sqrt{2}}{35} \right) . \]