I have a set of points, I want to select only a few of those points.
For that, I have 4 planes equations in the general form and I want to be able to check in a look if a given point would exist in between those 4 planes.
If you look at the image, the cyan and blue planes are parallel to each other, same for the yellow and red ones.
This question has been answered here: Determine if a point is within two planes
The idea is to find the distance between the 2 parallel planes, and for every point, if the distance of said point to one of the planes is greater than the distance between the 2 planes, I'd exlude that point (I'd check with the 2 other planes too, just an extra condition).
However, in that answer it is said to bring the equation into normal form. But all I can find to convert a plane equation from general form to normal form is this: Say the plane equation is $ax+by+cz+d=0$ Take the coeficients $a$, $b$ and $c$ to have: $\frac{ax+by+cz+d}{±\sqrt{a²+b²+c²}}=0$ The sign being the opposite of what the sign of d would be.
Then the distance to the origin would be $p= ± \frac{d}{\sqrt{a²+b²+c²}}$
Then in the answer it says to take a point in the second plane, but I don't know of any point in the planes themselves, but I guess I could calculate a random point by arbitrarly setting the $x$ and $y$ values and finding it's $z$ value.
I have no idea how to interpret $P_1 = \{x\in\mathbb R^3 | \langle x, \theta\rangle =s\}$
For ⟨⋅,⋅⟩ wich denotes the euclidean inner product okay I looked it up and understand what it does. But the rest I have no idea, it just looks to me like it's some kind of conditional probability.
Thx in advance for any
There's a simpler way to compute if a point $P=(x_P,y_P,z_P)$ lies between two parallel planes. As the planes are parallel, their equations can be put in the form $$ ax+by+cz=m \quad\text{and}\quad ax+by+cz=n. $$ Suppose for instance $m<n$. If: $$ m<ax_P+by_P+cz_P<n $$ then $P$ lies between the planes, otherwise it doesn't.