# Last vector-plane problem for the night

I would like to show that two planes are || and then find the distance between the two planes. I think I have the distance between the two planes covered pretty good, but I am not sure about the || part.

For example:

$$P_1$$ = 4x-2y+6z=3 and $$P_2$$ = -6x+3y-9z=4

then $$\vec{n}P_1$$ = <4,-2,6> and $$\vec{n}P_2$$ = <-6,3,-9>. So, my thinking is that if these two planes were || then they should have the same normal vector or at least a multiple of that normal vector. Is this correct, should I be looking at something else as well?

Office_Shredder
Staff Emeritus
Gold Member
P2 is: -3/2* (4x-2y+6z) = 4

Or 4x-2y+6z = -8/3

Does that help?

HallsofIvy
Homework Helper
prace said:
I would like to show that two planes are || and then find the distance between the two planes. I think I have the distance between the two planes covered pretty good, but I am not sure about the || part.

For example:

$$P_1$$ = 4x-2y+6z=3 and $$P_2$$ = -6x+3y-9z=4

then $$\vec{n}P_1$$ = <4,-2,6> and $$\vec{n}P_2$$ = <-6,3,-9>. So, my thinking is that if these two planes were || then they should have the same normal vector or at least a multiple of that normal vector. Is this correct, should I be looking at something else as well?
Yes, that's correct. And, just as obviously, as Office_Shredder showed, one is a multiple of the other. Now, to find the distance between them, pick any point in one plane, find the line through that point perpendicular to the plane (in the direction of the normal vector), find the point where that line intersects the other plane, and find the distance between the two points.