Q. 8

# 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).

Answer :

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

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