Question

Find the angle between the given planes. \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j}  + 2 \hat{k}  \right) = 6 \text{ and } \vec{r} \cdot \left( 3 \hat{i}  + 6 \hat{j}  - 2 \hat{k}  \right) = 9\]

Answer 1

` \text{ We know that the angle between the planes }  \vec{r} . \vec{n_1} = d_1 , \vec{r} . \vec{n_2} = d_2 \text{ is given by }`
\[\cos \theta = \frac{\vec{n_1} . \vec{n_2}}{\left| \vec{n_1} \right| \left| \vec{n_2} \right|}\]
\[ \text{ Here } , \vec{n_1} = 2 \hat{i} - \hat{j} + 2 \hat{k} ; \vec{n_2} = 3 \hat{i} + 6 \hat{j} - 2 \hat{k}  \]
\[\text{ So } ,\cos \theta = \frac{\left( 2 \hat{i}- \hat{j} + 2 \hat{k}  \right) . \left( 3 \hat{i}  + 6 \hat{j} - 2 \hat{k}  \right)}{\left| 2 \hat{i} - \hat{j} + 2 \hat{k} \right| \left| 3 \hat{i} + 6 \hat{j} - 2 \hat{k}  \right|} = \frac{6 - 6 - 4}{\sqrt{4 + 1 + 4} \sqrt{9 + 36 + 4}} = \frac{- 4}{\left( 3 \right) \left( 7 \right)} = \frac{- 4}{21}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{- 4}{21} \right)\]