# What is the distance between the planes 2x – 3y + 3z = 12 and –6x + 9y – 9z = 27?

Mar 11, 2016

Distance between the planes $2 x - 3 y + 3 z = 12$ and $- 6 x + 9 y - 9 z = 27$ is $4.48$.

#### Explanation:

First let us find a point on one plane. For this in plane $2 x - 3 y + 3 z = 12$, assume $x = y = 0$, then we have $3 z = 12$ or $z = 4$ and hence $\left(0 , 0 , 4\right)$ is on this plane.

Now we find the distance between point $\left(0 , 0 , 4\right)$ and plane $- 6 x + 9 y - 9 z = 27$ or $- 6 x + 9 y - 9 z - 27 = 0$.

As distance from a point $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ to plane $a x + b y + c z + d = 0$ is

$D = | a {x}_{1} + b {y}_{1} + c {z}_{1} + d \frac{|}{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}}$

The distance from point $\left(0 , 0 , 4\right)$ to plane $- 6 x + 9 y - 9 z - 27 = 0$
is given by

$D = | - 6 \times 0 + 9 \times 0 - 9 \times 4 - 27 \frac{|}{\sqrt{{6}^{2} + {9}^{2} + {9}^{2}}} = | - 36 - 27 \frac{|}{\sqrt{36 + 81 + 81}}$ or

$D = \frac{63}{\sqrt{198}} = \frac{63}{3 \sqrt{22}} = \frac{21}{\sqrt{22}} = 4.48$