Find the equations of planes parallel to the plane x – 2y + 2z = 3 which are at a unit distance from the point (1, 2, 3).
Formula : Plane = r . (n) = d
Where r = any random point
n = normal vector of plane
d = distance of plane from origin
If two planes are parallel , then their normal vectors are same
Therefore ,
Parallel Plane x – 2y + 2z – 3 = 0
Normal vector = (i - 2j + 2k)
∴ Normal vector of required plane = (i - 2j + 2k)
Equation of required planes r . (i - 2j + 2k) = d
In cartesian form x – 2y + 2y = d
It should be at unit distance from point (1,2,3)
Distance
For + sign = > 3 = 3 - d = > d = 0
For - sign = > 3 = - 3 + d = > d = 6
Therefore equations of planes are : -
For d = 0 For d = 6
x – 2y + 2y = d x – 2y + 2y = d
x – 2y + 2y = 0 x – 2y + 2y = 6
Required planes = x – 2y + 2y = 0
x – 2y + 2y – 6 = 0