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Question
Show that the normals to the following pairs of planes are perpendicular to each other.
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Solution
` \text{ Let } \vec{n_1} \text{ and } \vec{n_2} \text{ be the vectors which are normals to the planes } \vec{r} .\left( 2 \hat{i} - \hat{j} + 3 \hat{k} \right)= 5 \text{ and } \vec{r} .\left( 2 \hat{i} - 2 \hat{j} - 2 \hat{k} \right)= 5 \text{ respectively }. `
\[\text{ The given equations of the planes are }\]
\[ \vec{r} .\left( \text{ 2 }\hat{i} - \hat{j} + \text{ 3 }\hat{k} \right)= 5 ; \vec{r} .\left( \text{ 2 } \hat{i} - \text{ 2 } \hat{j} - \text{ 2 }\hat{k} \right)= 5\]
\[ \Rightarrow \vec{n_1} = \left( \text{ 2 }\hat{i} - \hat{j} + \text{ 3 }\hat{k} \right); \vec{n_2} = \left( \text{ 2 } \hat{i} - \text{ 2 } \hat{j} - \text{ 2 } \hat{k} \right)\]
\[Now, \vec{n_1} . \vec{n_2} = \left( \text{ 2 }\hat{i} - \hat{j} + \text{ 3 }\hat{k} \right) . \left( \text{ 2 }\hat{i} - \text{ 2 } \hat{j} - \text{ 2 }\hat{k} \right) = 4 + 2 - 6 = 0\]
\[\text{ So, the normals to the given planes are perpendicular to each other } .\]
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