Question

What is the distance between the planes `2x – 3y + 3z = 12` and `–6x + 9y – 9z = 27`?

Answer 1

Distance between the planes `2x-3y+3z=12` and `-6x+9y-9z=27` is `4.48`.

Explanation:

First let us find a point on one plane. For this in plane `2x-3y+3z=12`, assume `x=y=0`, then we have `3z=12` or `z=4` and hence `(0,0,4)` is on this plane.

Now we find the distance between point `(0,0,4)` and plane `-6x+9y-9z=27` or `-6x+9y-9z-27=0`.

As distance from a point `(x_1,y_1,z_1)` to plane `ax+by+cz+d=0` is

`D=|ax_1+by_1+cz_1+d|/sqrt(a^2+b^2+c^2)`

The distance from point `(0,0,4)` to plane `-6x+9y-9z-27=0`
is given by

`D=|-6xx0+9xx0-9xx4-27|/sqrt(6^2+9^2+9^2)=|-36-27|/sqrt(36+81+81)` or

`D=63/sqrt(198)=63/(3sqrt22)=21/sqrt22=4.48`