## Distance Between Two Parallel Plans

Hey all,
I was working out of my textbook over the summer preparing for Calculus III, and I have a question about finding the distance between two parallel planes.

I believe that I understand how the equation given works:
$$D= \frac{| ax_{o} + by_{o} + cz_{o} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$$

To find the point, I set z=y=0. However, assuming that the planes are parallel, why can I not use this to find the point (x,0,0) on both planes and calculate the distance between these two points? It seems to me that since both points are in their respective planes, the distance given would work, but I get a difference answer from the above formula.

Why is that?
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 Quote by Erikh86 Hey all, I was working out of my textbook over the summer preparing for Calculus III, and I have a question about finding the distance between two parallel planes. I believe that I understand how the equation given works: $$D= \frac{| ax_{o} + by_{o} + cz_{o} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$$ To find the point, I set z=y=0. However, assuming that the planes are parallel, why can I not use this to find the point (x,0,0) on both planes and calculate the distance between these two points? It seems to me that since both points are in their respective planes, the distance given would work, but I get a difference answer from the above formula. Why is that?
Do you mean just finding an arbitrary point from each plane and the calculating the distance between those two points? Or are you suggesting that there is a point (x,0,0) that is on both planes?

If it is the first case, then if you did this, you would not be getting the SHORTEST distance from one plane to the other, which is what you want. That is why you have to project a vector going from Plane1 to Plane2 onto Plane1. The length of this projection is the distance.
 Hey Erikh86 and welcome to the forums. The easiest way IMO is to just compare the two distance coeffecients (i.e. the values of d) for both planes. If your normals have the same direction (and not opposite), then if you plug in the zero vector for your plane equations, you will get a d value for both plane equations. Then simple compare the magnitude of the difference and you should get the distance between the two planes (i.e. |d1 - d2|). These d values correspond to information regarding an existing point on the plane that takes into account the normal vector, but the method should give you the distance as long as the unit normals of the plane itself are equal.

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