Q. 125.0( 1 Vote )

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.

Answer :

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





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