Question

Find the angle between the given planes.

\[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j}  - 6 \hat{k}  \right) = 5 \text{ and } \vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right) = 9\]

 

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 }`

\[\cos \theta = \frac{\vec{n_1} . \vec{n_2}}{\left| \vec{n_1} \right| \left| \vec{n_2} \right|}\]

\[ \text{ Here } , \vec{n_1} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} ; \vec{n_2} = \hat{i} - 2 \hat{j}  + 2 \hat{k}  \]

\[ \text{ So } ,\cos \theta = \frac{\left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k} \right) . \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right)}{\left| 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right| \left| \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right|} = \frac{2 - 6 - 12}{\sqrt{4 + 9 + 36} \sqrt{1 + 4 + 4}} = \frac{- 16}{\left( 7 \right) \left( 3 \right)} = \frac{- 16}{21}\]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{- 16}{21} \right)\]