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

##### 1 Answer

#### 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:

The equation of the plane parallel to the original one passing through

#vec n*"< "x-x_0,y-y_0,z-z_0> =0#

#<1,2,-2>*"<"x-x_0,y-y_0,z-z_0> =0#

#x-x_0+2y-2y_0-2z+2z_0=0#

#x+2y-2z-x_0-2y_0+2z_0=0#

Or

#x+2y-2z+d=0# [1]

where#a=1# ,#b=2# ,#c=-2# and#d=-x_0-2y_0+2z_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:

Distance between 2 parallel planes

In the original plane let's choose a point.

For instance, when

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=|ax_1+by_1+cz_1+d|/sqrt(a^2 + b^2 + c^2)#

#2=|1*0+2*0+(-2)*(-1/2)+d|/sqrt(1+4+4)#

#|d+1|=2*3# =>#|d+1|=6# First solution:

#d+1=6# =>#d=5#

#-> x+2y-2z+5=0# Second solution:

#d+1=-6# =>#d=-7#

#-> x+2y-2z-7=0#