# What are the equations of the planes that are parallel to the plane x+2y-2z=1 and two units away from it?

Feb 15, 2016

$x + 2 y - 2 z + 5 = 0$ and $x + 2 y - 7 = 0$

#### Explanation:

First we'll find the equation of ALL planes parallel to the original one.
As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
$\vec{n} = < 1 , 2 - 2 >$

The equation of the plane parallel to the original one passing through $P \left({x}_{0} , {y}_{0} , {z}_{0}\right)$ is:

$\vec{n} \cdot \text{< } x - {x}_{0} , y - {y}_{0} , z - {z}_{0} > = 0$
$< 1 , 2 , - 2 > \cdot \text{<} x - {x}_{0} , y - {y}_{0} , z - {z}_{0} > = 0$
$x - {x}_{0} + 2 y - 2 {y}_{0} - 2 z + 2 {z}_{0} = 0$
$x + 2 y - 2 z - {x}_{0} - 2 {y}_{0} + 2 {z}_{0} = 0$

Or

$x + 2 y - 2 z + d = 0$ [1]
where $a = 1$, $b = 2$, $c = - 2$ and $d = - {x}_{0} - 2 {y}_{0} + 2 {z}_{0}$

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

In the original plane let's choose a point.
For instance, when $x = 0$ and $y = 0$:
$x + 2 y - 2 z = 1$ => $0 + 2 \cdot 0 - 2 z = 1$ => $z = - \frac{1}{2}$
$\to {P}_{1} \left(0 , 0 , - \frac{1}{2}\right)$

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping $D = 2$, and $d$ as $d$ itself, we get:

$D = | a {x}_{1} + b {y}_{1} + c {z}_{1} + d \frac{|}{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}}$
$2 = | 1 \cdot 0 + 2 \cdot 0 + \left(- 2\right) \cdot \left(- \frac{1}{2}\right) + d \frac{|}{\sqrt{1 + 4 + 4}}$
$| d + 1 | = 2 \cdot 3$ => $| d + 1 | = 6$

First solution:
$d + 1 = 6$ => $d = 5$
$\to x + 2 y - 2 z + 5 = 0$

Second solution:
$d + 1 = - 6$ => $d = - 7$
$\to x + 2 y - 2 z - 7 = 0$