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

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

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

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

$=1/6$

What is the distances between the parallel planes 2x-2y+z+3=02x-2y+z+3=0 and 4x-4y+2z+5=04x-4y+2z+5=0?