Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?

Lets Find It Out..

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:
→n=<1,2−2>

The equation of the plane parallel to the original one passing through P(x0,y0,z0)is:

→n⋅< x−x0,y−y0,z−z0>=0
<1,2,−2>⋅<x−x0,y−y0,z−z0>=0
x−x0+2y−2y0−2z+2z0=0
x+2y−2z−x0−2y0+2z0=0

Or

x+2y−2z+d=0 [1]
where a=1, b=2, c=−2 and d=−x0−2y0+2z0

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:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when x=0 and y=0:
x+2y−2z=1 => 0+2⋅0−2z=1 => z=−12
→P1(0,0,−12)

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:

well the answer to that is xoxoxo

step-by-step explanation:

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.