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.

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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