Distance between two planes

  • Thread starter james_j_j
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  • #1
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Homework Statement


Find the distance between the planes 4x + 5y - 3z + 4 = 0, and -8x -10y + 6z + 12 = 0.


Homework Equations



V = (x, y, z) + Y(a, b, c)


The Attempt at a Solution



Okay, this problem has been driving me insane. Even though it's incredibly simple, and I know there's an easy formula, I haven't been able to do it the way the lecturer showed us.

Plane 1): 4x + 5y - 3z = -4
Plane 2): -8x - 10y + 6z = -12

Point on 1) = P = (-1, 0, 0)

Let v be the vector connecting P and Q (point on P2).

v = (-1, 0, 0) + x(4, 5, -3)

Intersect v and P2.

(-1, 0, 0) + x(4, 5, -3) = -12
-1 + 4x + 5x -3x = -12
6x = -11
x = -11/6

-> Q = (-1, 0, 0) - 11/6 (4, 5, -3)

Here's the problem, "Q" doesn't seem to be on P2, which was the entire point of solving for x.

e.g. sub x,y,z = (-1 - 44/6), (-55/6), (-33/6) into P2 and you should get -12 (as I understand it), but you get ~125.

Any help appreciated.
 

Answers and Replies

  • #2
Your problem is quite simple but the I way I would solve it requires you to know about vector projection, are you familiar with it?
 
  • #3
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Yes, we've covered vector projection.
 
  • #4
ehild
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Two planes are either parallel or they intersect. Parallel planes have common normal vector n. Are these planes parallel? The equation of a plane is (r-r0)n=0, or rn=r0n where r is the position vector of an arbitrary point on the plane and r0 is that of a fixed point on it. Using unit normal vector, you get at once how far is a plane from the origin. Think how to get the projection of a vector to an other one, using dot product. Make a figure and you will see.

This is the easy way. Yours is more complicated. And what you did to find the intercept of that line with P2 is totally wrong.

"Intersect v and P2.

(-1, 0, 0) + x(4, 5, -3) = -12
-1 + 4x + 5x -3x = -12
6x = -11
x = -11/6"


You made a vector equal to a scalar. Check it.

ehild
 
  • #5
Ok, keep in mind that the general equation of a plane is [tex]Ax+By+Cz+D=0 [/tex].
It can be proven that the vector [tex]\bf{N}=(A,B,C) [/tex], called normal vector, is orthogonal to the plane [tex] Ax+By+Cz+D=0 [/tex]. So two parallel planes have a common normal vector [tex] \bf{N}[/tex].

What do you need to be able to talk about the "distance" between two planes? What would happen if they intersected each other somewhere? what would that distance be?

Now, take a point [tex] x_0 [/tex] in plane 1 and take another point [tex] x_1 [/tex] in plane 2, and project vector [tex] x_0 - x_1 [/tex] over vector [tex] \bf{N} [/tex]. The length of said projection should be your distance.
 
  • #6
HallsofIvy
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The only way two planes have a distance between them is when they are parallel- and these planes are because the coefficients of x, y, z in one equation are a constant multiple of the corresponding coefficients in the other. And finding the distance does not necessarily require "vector projection".

Choose any point in one plane- choose x and y to be whatever you like and solve the equation of the plane for z.

Find the line through that point perpendicular to the plane- that's easy. For example, if the point is [itex](x_0, y_0, z_0)[/itex] then the line through that point perpendicular to Ax + By + Cz + C = 0 has parametric equations [itex]x= At+ x_0[/itex], [itex]y= Bt+ y_0[/itex], [itex]z= Ct+ z_0[/itex].

Find the point where that line intersects the other plane. Since the planes are parallel, the line is also perpendicular to the second plane and the distance between those two points is the distance between the two planes.
 

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