Question

Find the equations of the planes parallel to the plane x − 2y + 2z − 3 = 0 and which are at a unit distance from the point (1, 1, 1).

 

Answer 1

\[\text{ The equation of the plane parallel to the given plane is } \]
\[x - 2y + 2z + k = 0 . . . \left( 1 \right)\]
\[ \text{ It is given that plane (1) is at a distance of 1 unit from (1, 1, 1). } \]
\[ \Rightarrow \frac{\left| 1 - 2 + 2 + k \right|}{\sqrt{1^2 + \left( - 2 \right)^2 + 2^2}} = 1\]
\[ \Rightarrow \frac{\left| 1 + k \right|}{3} = 1\]
\[ \Rightarrow \left| 1 + k \right| = 3\]
\[ \Rightarrow 1 + k = 3; 1 + k = - 3\]
\[ \Rightarrow k = 2; k = - 4\]
\[ \text { Substituting these two values one by one in (1), we get } \]
\[x - 2y + 2z + 2 = 0 \text{ and} \]
\[x - 2y + 2z - 4 = 0, \text{ which are the equations of the required planes} .\]