Question

I'm having trouble with this one.

Problem: Find the equations for the planes parallel to x+2y-2z=1 and two units away from it.

Answer 1

If a point (p,q,r) is not on the place Ax+By+Cz+D = 0 then the distance to the plane is

(Ap + Bq + Cr + D)/(sqrt(A^2+B^2+C^2))

So let the new equation be x + 2y - 2z + D = 0
and pick any point which satisfies the given equation, say (1,0,0)

So the distance is (1 + 4 + 4 + C)/sqrt(1+0+0)
= (1+D)/sqrt(5) but this is supposed to be 2

(1+D)/sqrt(5) = 2
D = 2SQRT(5) - 1

and your new equation is
x + 2y - 2z + 2SQRT(5) - 1 = 0

Answer 2

I put the 4's in the wrong place, it should have said

"So the distance is (1 + 0 + 0 + C)/sqrt(1+4+4)
= (1+D)/sqrt(5) but this is supposed to be 2 "

the rest of the post is ok