# General form of plane equation

1. Jul 23, 2010

### p1ayaone1

...is Ax + By + Cz + D = 0.

The vector <A, B, C> is a normal vector of the plane. My question is: does the value of D have any geometric significance/interpretation?

I have an algorithm (that I didn't write myself) to evaluate the best-fit plane for a set of points in R3, and the value of D is coming back extremely large (10^20 or something obviously ridiculous). I wonder if D is just a mathematical artifact and I shouldn't worry, or if there is a problem with the algorithm (or my usage of it).

I don't think D should be that large based on the equation for distance between a point (x0, y0, z0) and a plane Ax + By + Cz + D, which is

D = abs(A*x0 + B*y0 + C*z0+ D) / sqrt(A^2 + B^2 + C^2)

My values of A, B, and C are all -1<value<1, but D is so big that it will completely dominate that expression.

Maybe this is a junior question and this thread should be re-categorized as such.

Thanks

2. Jul 23, 2010

### Anthony

The equation of a plane with normal $$\mathbf{n}$$ going through the point $$\mathbf{a}$$ is given by $$(\mathbf{x} - \mathbf{a})\cdot \mathbf{n}=0$$. Can you see how that helps?

3. Jul 23, 2010

### p1ayaone1

sure. Expanding the dot product gives the point-normal form (as opposed to the general form) of the plane equation. If a = (x0, y0, z0), x = <x, y, z> and n = <A, B, C>, we have A(x-x0) + B(y-y0) + C(z-z0) = 0

That means D = -(x0 + y0 + z0), but I still don't see what the significance of that quantity is...

4. Jul 24, 2010

### Anthony

Well, using your notation we have:

$$(\mathbf{x} - \mathbf{a})\cdot \mathbf{n} =0 \quad \textrm{means} \quad Ax+By+Cz = \mathbf{n}\cdot \mathbf{a}$$

So your D is given by:

$$D = - \mathbf{n}\cdot \mathbf{a} = -|\mathbf{a}| \cos \theta = |\mathbf{a}| \cos (\theta - \pi)$$

where $$\theta$$ is the angle made between the vectors $$\mathbf{a}$$ and $$\mathbf{n}$$. So you have a geometrical interpretation of D.