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

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Answer 1 · 2017-07-11T05:36:30

The distance is $= \frac{1}{6}$ $=1/6$

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

Then,

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

$d = \frac{| a x_{1} + b y_{1} + c z_{1} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$ $d=|ax_1+by_1+cz_1+d|/sqrt(a^2+b^2+c^2)$

$= \frac{| 4 \cdot 0 - 4 \cdot 0 - 2 \cdot 3 + 5 |}{\sqrt{16 + 16 + 4}}$ $=|4*0-4*0-2*3+5|/sqrt(16+16+4)$

$= \frac{| - 1 |}{\sqrt{36}}$ $=|-1|/sqrt(36)$

$= \frac{1}{6}$ $=1/6$

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?