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Question
Show that the normals to the following pairs of planes are perpendicular to each other.
x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0
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Solution
` \text{ Let } \vec{n_1} \text{ and } \vec{n_2} \text{ be the vectors which are normals to the planesx - y + z = 2 and 3x + 2y - z = - 4 respectively }.`
\[\text{ The given equations of the planes are } \]
\[x - y + z = 2; 3x + 2y - z = - 4\]
\[ \Rightarrow \left( x \hat{i} + y \hat{j} + z \hat{k} \right) . \left( \hat{i} - \hat{j} + \hat{k} \right) = 8; \left( x \hat{i} + y \hat{j} + z \hat{k} \right) . \left( 3 \hat{i} + 2 \hat{j} - \hat{k} \right) = - 4\]
\[ \Rightarrow \vec{n_1} = \hat{i} - \hat{j} + \hat{k} \text{ ; } \vec{n_2} = \text{ 3 }\hat{i} + \text{ 2 }\hat{j} - \hat{k} \]
\[\text{ Now } , \vec{n_1} . \vec{n_2} = \left( \hat{i} - \hat{j} + \hat{k} \right) . \left( \text{ 3 }\hat{i} + \text{ 2 }\hat{j} - \hat{k} \right) = 3 - 2 - 1 = 0\]
\[ \text{ So, the normals to the given planes are perpendicular to each other } .\]
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