Planar Inequalities

May 2015
2
0
USA
My goal here is to figure out a routine to determine if a point lies within a certain distance of a plane.

I understand the standard form of planes (ax+by+cz=d) and that the equation itself represents the normal vector, but what I'm having trouble with is d. I know you can test a point on a plane by plugging in the values and seeing if the distance values match. So then, to see if the point is less than a given distance away, would it be as simple as measuring the change in d?

Thanks for your help.
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey BluEyeDevill.

The standard distance formula for a point with respect to a plane is simply d = |<n,p-p0>| where p is the point you are testing and p0 is an existing point on the plane.

The plane equation is always ax + by + cz + d = 0 but when there is an excess distance you will get ax + by + cz + d = e where |e| represents a distance from the plane itself.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Equivalently, let the plane be given by Ax+ By+ Cz= D and the point be \(\displaystyle (x_0, y_0, z_0)\).

The "distance from point P to plane q" is, by definition, the shortest distance from P to any point in the plane.
And that must be along a line perpendicular to the plane- to see that choose any other point in the plane and draw the lines from P to those two points.
One, the perpendicular, is a leg of a right triangle while the other is a hypotenuse- and the hypotenuse of a right triangle is always longer than a leg.

A vector perpendicular to Ax+ By+ Cz= D is of the form <A, B, C> so a line through \(\displaystyle (x_0, y_0, z_0)\) in the direction of that line is \(\displaystyle x= At+ x_0\), \(\displaystyle y= Bt+ y_0\), and \(\displaystyle z= Ct+ z_0\). That line crosses the plane when A(At+ x_0)+ B(Bt+ y_0)+ C(Ct+ z_0)= D or \(\displaystyle (A^2+ B^2+ C^2)t+ (Ax_0+ By_0+ Cz_0)= D\)
so \(\displaystyle t= \frac{D- Ax_0- By_0- Cz_0}{A^2+ B^2+ C^2}\).

Put that value of t into the equation of the plane to find the point where the line crosses the plane and then find the distance between that point and \(\displaystyle (x_0, y_0, z_0)\).
 

Plato

MHF Helper
Aug 2006
22,507
8,664
My goal here is to figure out a routine to determine if a point lies within a certain distance of a plane.
Equivalently, let the plane be given by $\Pi:~ Ax+ By+ Cz= D$, the normal is \(\displaystyle N={A\bf{i}}+{B\bf{j}}+{C\bf{k}}\).
Suppose that $Q\in\Pi$, then "distance from point $P$ to plane $\Pi$" is, $\dfrac{| \overrightarrow {PQ}\cdot N |}{\|N\|}$
 
May 2015
2
0
USA
I never thought about it that way. So its essentially a vector projection?

Thanks for the help!

Over the past few days I realized that when the equation is equal to 0 you have a plane that is simply located at the origin, and if you multiply the equation by a scalar quantity then d would represent the distance to the origin, or the vector magnitude. so then, yes, (along the lines of what you said) the change in d can be directly tested because it represents the shortest distance to the plane.

Awesome! Is there a way I can mark my thread solved?