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Question
Find the equation of the plane through (3, 4, −1) which is parallel to the plane \[\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + 2 = 0 .\]
Solution
\[\text{ Let the equation of a plane parallel to the given plane be } \]
\[ \vec{r} . \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) = k . . . \left( 1 \right)\]
\[\left( x \hat{i} + y \hat{j} + z \hat{k} \right) . \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) = k \]
\[ \text{ This passes through (3, 4, -1) .So } ,\]
\[\left( 3 \hat{i} + 4 \hat{j} - \hat{k} \right) . \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) = k\]
\[ \Rightarrow k = 6 - 12 - 5 = - 11\]
\[\text{ Substituting this in (1), we get } \]
\[ \vec{r} . \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) = - 11\]
\[ \vec{r} . \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + 11 = 0, \text{ which is the equation of the required plane} .\]
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