What is the distances between the parallel planes #2x-2y+z+3=0# and #4x-4y+2z+5=0#?

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Answer 1 · 2017-07-11T05:36:30

The distance is #=1/6#

We need a point on the first plane #P=(0,0,-3)#

Then,

The distance between #P# and #4x-4y+2z+5=0# is

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

#=|4*0-4*0-2*3+5|/sqrt(16+16+4)#

#=|-1|/sqrt(36)#

#=1/6#