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

1 Answer
Shwetank Mauria
Mar 11, 2016

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#

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