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arrow_back The planes $$ x+2 y-2 z=3 \text { and } 2 x+4 y-4 z=7 $$ are parallel since their normals, $(1,2,-2)$ and $(2,4,-4)$, are parallel vectors. Find the distance between these planes.

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The planes $$ x+2 y-2 z=3 \text { and } 2 x+4 y-4 z=7 $$ are parallel since their normals, $(1,2,-2)$ and $(2,4,-4)$, are parallel vectors. Find the distance between these planes.

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To find the distance $D$ between the planes, we can select an arbitrary point in one of the planes and compute its distance to the other plane.

By setting $y=z=0$ in the equation $x+2 y-2 z=3$, we obtain the point $P_{0}(3,0,0)$ in this plane.

The distance between $P_{0}$ and the plane $2 x+4 y-4 z=7$ is $$ D=\frac{|2(3)+4(0)+(-4)(0)-7|}{\sqrt{2^{2}+4^{2}+(-4)^{2}}}=\frac{1}{6} $$
by Platinum
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