Question

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

Answer 1

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

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

\[\text{ Let }  \vec{b}  \text { be a vector parallel to the vector with direction ratios }1, 2, 2 . \]

\[ \Rightarrow \vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k} \]

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

\[\text{ Now, }\]

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

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

\[ = \frac{2 + 6 + 12}{\sqrt{4 + 9 + 36} \sqrt{1 + 4 + 4}}\]

\[ = \frac{20}{21} \]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{20}{21} \right)\]