# Planar Inequalities

#### BluEyeDevill

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?

#### chiro

MHF Helper
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
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
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, $\dfrac{| \overrightarrow {PQ}\cdot N |}{\|N\|}$

#### BluEyeDevill

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?