Asked by Joseph
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.
Problem: Find the equations for the planes parallel to x+2y-2z=1 and two units away from it.
Answers
Answered by
Reiny
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
(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
Answered by
Reiny
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
"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