# Distance between point-plane & plane-plane

1. Nov 3, 2008

Greetings,

If I have two planes,

2x - y + 5z = 4 and -4x + 2y - 10z = 5

can I give the distance between them by simply subtracting their points?
(8x,-3y-5z) ?

(I intended to set this up so that they do not intersect, I hope I did that correctly).

Furthermore, can anyone shed some light on why the distance formula for a point - plane is

$$D= \frac{| ax_{o} + by_{o} + cz_{o} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$$

where a,b,c,d are points on the plane and x,y,z are the point coordinates?

Thank you!

2. Nov 3, 2008

### mgb_phys

3. Nov 3, 2008

Thanks! "Easy" meaning, what I did is correct? Thanks for that article also.

4. Nov 3, 2008

### mgb_phys

Since any point on one of theplanes must satisfy the plane equation you could simply find a point on that plane and then use the point-plane equation you find in the second part

5. Nov 4, 2008

### HallsofIvy

Obviously, that depends on x, y, and z and is not "the" distance. (Oh, and 2-(-4)= 6, not 8.) As has been said, two planes do not intersect if and only if they are parallel but choosing any point in one plane, there exist an infinite number of distance from that point to points in the other plane. "The" distance from one object to another is always defined as the minimum distance (or infimum) from points in one to the other. It is easy to see that the minimum distance from one plane to a parallel plane is along the perpendicular (use the Pythagorean theorem). In your example, since the vector <-4, 2, -10>= -2<2, -1, 5> is normal to the plane we can get a perpendicular line from it.
1) Choose any point in the first plane. If I take z= 0, x= 1, then 2- y= 4 or y= -2. (1, -2, 0) is a point in the plane.

2) Write the equation of the line through that point having normal vector to the plane as direction vector: x= 1+ 2t, y= -2- t, z= 5t are the parametric equations of a line through (1, -2, 0) perpendicular to the plane.

3) Find where that line intersects the second plane. That is, replace x, y, and z in the second equation by those parametric values: (-4)(1+ 2t)+ 2(-2-t)-10(5t)= 5. That is, -4- 8t- 4- 2t- 50t= -60t- 8= 5 so -60t= 13 and t= -(13/60). The point at which the line, through (1, -2, 0), perpendicular to the first plane, intersects the second plane is given by putting t= -13/60 in the parametric equations. Once you have that second point, calculate the distance between the two points.

Again, by "distance between a point and a plane" we mean the shortest distance from that point to any point in the plane- and again that is along the line perpendicular to the plane. I will assume that you mean the point is given by (x0, y0, z0) and the plane is ax+ by+ cz+ d= 0.
Then <a, b, c> is a vector perpendicular to the plane and a line through (x0, y0, z0) perpendicular to the plane is given by x= x0+ at, y= y0+ bt, z= z0+ ct. That intersects the plane when a(x0+at)+ b(y0+ bt)+ c(z0+ ct)+ d= 0.

Multiplying that out, ax0+ a2t+ by0+ b2t+ cz0+ c2t= 0 so (a2+ b2+ c2
)t= -(ax0+ by0+ cz0+ d).

Now solve for t, put it into the parametric equations to find the point at which the line intersects the plane and find the distance between those two points.

6. Nov 4, 2008

Thanks a lot HallsofIvy, sorry If I'm being a bit slow here...

At this point, any line can be represented by those right? Not just those perpendicular to the plane...

This is because the dot product of two orthogonal vectors is 0, yes?

7. Nov 5, 2008

### HallsofIvy

At any point a line can be represented by x= x0+ at, y= y0+ bt, z= z0+ ct. That line is perpendicular to the plane if and only if <a, b, c> is the normal to the plane which is the same as saying the plane is given by ax+ by+ cz= constant.

That is because you can put the x, y, z, values of the point into the equation of the plane.