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#
