Question

Find the angle between the given planes. \[\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) = 1 \text{ and } \vec{r} \cdot \left( - \hat{i}  + \hat{j}  \right) = 4\]

 

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 }`

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

\[Here, \vec{n_1} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}    ;  \vec{n_2} = - \hat{i} + \hat{j} + 0 \hat{k} \]

\[\text{ So } ,\cos \theta = \frac{\left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k}  \right) . \left( - \hat{i}  + \hat{j}  + 0 \hat{k} \right)}{\left| 2 \hat{i}  - 3 \hat{j}  + 4 \hat{k}  \right| \left| - \hat{i} + \hat{j} + 0 \hat{k}  \right|} = \frac{- 2 - 3}{\sqrt{4 + 9 + 16} \sqrt{1 + 1 + 0}} = \frac{- 5}{\sqrt{29} \sqrt{2}} = \frac{- 5}{\sqrt{58}}\]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{- 5}{\sqrt{58}} \right)\]