Find the equation of the plane parallel to the plane 2x – 3y + 5z + 7 = 0 and passing through the point (3, 4, - 1). Also, find the distance between the two planes.
Formula : Plane = r . (n) = d
Where r = any random point
n = normal vector of plane
d = distance of plane from origin
The distance between two parallel planes, say
Plane 1:ax + by + cz + d1 = 0 &
Palne 2:ax + by + cz + d2 = 0 is given by the formula
If two planes are parallel , then their normal vectors are same
Therefore ,
Parallel Plane 2x – 3y + 5z + 7 = 0
Normal vector = (2i - 3j + 5k)
∴ Normal vector of required plane = (2i - 3j + 5k)
Equation of required plane r . (2i - 3j + 5k) = d
In cartesian form 2x – 3y + 5y = d
Plane passes through point (3,4, - 1) therefore it will satisfy it.
2(3) – 3(4) + 5( - 1) = d
6 – 12 - 5 = d
d = - 11
Equation of required plane 2x – 3y + 5z = - 11
2x – 3y + 5z + 11 = 0
Therefore ,
First Plane 2x – 3y + 5z + 7 = 0 …… (1)
Second plane 2x – 3y + 5z + 11 = 0 …… (2)
Using equation (1) and (2)
Distance between both planes