What is the formula for finding the distance between two parallel planes?

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In summary, to find the distance between two planes, you first need to find the normal vector, which is perpendicular to the planes, and project it onto the plane. The line that intersects the plane at that point is the distance between the planes.
  • #1
james_j_j
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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.
 
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  • #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
Yes, we've covered vector projection.
 
  • #4
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
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.
 

1. What is the formula for finding the distance between two planes?

The formula for finding the distance between two planes is:d = |ax1 + by1 + cz1 - ax2 - by2 - cz2| / √(a2 + b2 + c2)

where (x1, y1, z1) and (x2, y2, z2) are points on the two planes and (a, b, c) is the normal vector of the planes.

2. Can the distance between two planes be negative?

No, the distance between two planes is always a positive value. It represents the shortest distance between the two planes.

3. What is the significance of finding the distance between two planes?

Finding the distance between two planes is useful in various applications such as 3D geometry, computer graphics, and physics. It can help determine the relationship between two planes and the shortest distance between them.

4. How is the distance between two parallel planes calculated?

If the two planes are parallel, the distance between them is equal to the distance of a point on one plane to the other plane. This can be calculated using the formula: d = |ax + by + cz - d| / √(a2 + b2 + c2), where (x, y, z) is a point on one of the planes and d is the distance of that point to the other plane.

5. Is there a visual representation of the distance between two planes?

Yes, the distance between two planes can be visualized as the distance between two parallel lines that are perpendicular to both planes. This can be seen in 3D space or on a 2D coordinate plane.

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