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Calculating the Distance Between a Point and a Plane

Date: 10/12/95 at 19:57:12
From: Anonymous
Subject: Distance between a point and a plane

I am trying to calculate the minimum distance between 
a point, located in 3-dimensional space, and a plane.
I define the plane by determining the equation 
Ax + By + Cz + D = 0, from three points, each with
x,y,z coordinates.( I think I can do this!)
Next I want to determine the minimum distance from this 
plane to a point which has x,y,z coordinates.  I believe
the minimum distance from the plane to the point is the 
perpendicular distance from the plane to the point.
I am unsure of the equations to perform this.

Date: 10/15/95 at 10:50:45
From: Doctor Ethan
Subject: Re: Distance between a point and a plane

The equation is: Ax + By + Cz + D
               \/ A^2 + B^2 + C^2

In this equation the x,y,z are the point in question and the A,B,C,D
are the values in the standard expression for the plane.
It is a lot like the equation for the distance from a line.

Now we can figure out how to derive it.

Let us sart by finding a vector normal to the plane.  
Certainly the most convienient is {A,B,C}.

Did you know that it will always be normal to the plane?

I will leave it to you to check that truth.

Now the derivation is pretty simple.pick any point on the plane, examine the
vector between it and the desired point.

Let's call the point in question P and the point we have choen on the plane
D.  Then we can make a side view drawing that looks like this.

    * *
   *  *
  *   *
 *    *

Now what we are look ing for is that veritcal distance.  One way to find it,
would be to find that bottom point and then use the Dot Product to find the
distance between them.

However we can use that Normal vector to do something slicker.  I am going to
make a bigger drawing to help you get the picture.
In this drawing I an going to but the tail of the vector {A,B,C} at point D.

And I will call its head N.

So we have...

        * *
       *  *
N     *   *
*    *    *
*   *     *
*  *      *
* *       *
**        *
D*********O (Call it O just so we can talk about it)

Now again since we know the points D and P, and  we know the vector {A,B,C},
we can do a neat trick.

We know from Geometry that 

Length of PO
------------   =  Cos (Angle P)
Length of DP

Which is 

Length of PO = (Length of DP) Cos (Angle P)
[I will refer to this later as equation 1]

But the Cos (Angle P) equals Cos (Angle NDP)

And you may recall that the Dot product of two vectors is also their lengths
times the Cos of the angle in between them.

So {A,B,C} . (P-D)  =  Length of {A,B,C} * Length of DP * Cos (Angle NDP)

That is pretty close to what we had in equation 1.

All we have to do is divide by length of {A,B,C}

So we have this  

               {A,B,C} . (P-D)
Length of PO = ---------------
               Length of {A,B,C}

Now if we choose P to be {x,y,z}  and let D be {0,0,-D/C}

You should see how we got the above answer.

Hope that helps......

-Doctor Ethan,  The Geometry Forum

Associated Topics:
College Linear Algebra

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