Find the distance between the parallel planes 2x + 3y + 4 = 4 and 4x + 6y + 8z = 12.
Formula : The distance between two parallel planes, say
Plane 1:ax + by + cz + d1 = 0 &
Plane 2:ax + by + cz + d2 = 0 is given by the formula
where () are constants of the planes
Therefore ,
First Plane 2x + 3y + 4 = 4
2x + 3y + 4 – 4 = 0 …… (1)
Second plane 4x + 6y + 8z = 12
4x + 6y + 8z - 12 = 0
2(2x + 3y + 4z – 6) = 0
2x + 3y + 4z – 6 = 0 …… (2)
Using equation (1) and (2)
Distance between both planes