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

Question

Want to understand the work involved

Answer 1

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

The normal vector n, to the plane is also the directional vector of a line perpendicular to it. So we will pick a point on the plane and find the two points a distance 2 away in either direction.

n = <1, 2, -2>

|| n || = √[1² + 2² + (-2)²] = √(1 + 4 + 4) = √9 = 3

To get a direcitonal vector of magnitude 2, multiply by 2/3.

n = (2/3)<1, 2, -2> = <2/3, 4/3, -4/3>

Now pick a point P, on the plane. Let y = z = 0. Then x = 1.
P = P(1, 0, 0).

Planes parallel to the given plane will have the same coefficients for each of the variables. Only the constant will change. Find the points Q and R on the two parallel planes.

Q = P + n = <1, 0, 0> + <2/3, 4/3, -4/3> = <5/3, 4/3, -4/3>
R = P - n = <1, 0, 0> - <2/3, 4/3, -4/3> = <1/3, -4/3, 4/3>

Plug in the points to find the equations of the two parallel planes.

Q:
x + 2y - 2z = 5/3 + 2(4/3) - 2(-4/3) = 21/7 = 7
x + 2y - 2z = 7

R:
x + 2y - 2z = 1/3 + 2(-4/3) - 2(4/3) = 21/7 = -15/3 = -5
x + 2y - 2z = -5

The two parallel planes a distance 2 away from the given plane are:

x + 2y - 2z = 7
x + 2y - 2z = -5

Answer 2

x-1