Distance between two parallel planes 2x+y+2z=8 and 4x+2y+4z+5=0 is

Question

Distance between two parallel planes $2x+y+2z=8$ and $4x+2y+4z+5=0$ is

Accepted Answer

The planes are $$2x + y + 2x - 8 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $4x + 2y + 4z + 5 = 0$

or $$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\therefore$$ Distance between $$\left( 1 \right)$$ and $$\,\left( 2 \right)$$

$$ = \left| {{{{5 \over 2} + 8} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$$

$$ = \left| {{{21} \over {2\sqrt 9 }}} \right|$$

$$ = {7 \over 2}$$

Solutions · 1