I have a line $L_1$ defined as $P_1 = (x_1, y_1, z_1)$ and $P_2 = (x_2, y_2, z_2)$. I wish to find a point(s) that is perpendicular to the line about its end $P_2$ (could be generalized to any given point along the line) that is a known normalized distance away of length l on that plane ($\perp$ to $L_1$).
I realized plotting enough points you eventually get what would look like a circle about the line, I'm only interested in plotting n number of points each equal angles apart on the plane, doesn’t matter how the first point is started as long as the above criteria is met.
Image link below I've used $n=5$ for example
3D CAD ILLUSTRATION OF WHAT IM TRYING TO ACHIEVE
Any help is much appreciated (bonus if know how to implement in python...but not necessary :))
Let $(a,b,c):=(x_2-x_1,y_2-y_1,z_2-z_1)$ and $(\alpha,\beta,\gamma):=N:=\frac1{\sqrt{a^2+b^2+c^2}}(a,b,c).$
Choose some vector $V:=(u,v,w)$ of length $l,$ and orthogonal to $N$ i.e. such that $au+bv+cw=0.$ For instance, if $a$ or $b$ is $\ne0$: $$u:=\frac{bl}{\sqrt{a^2+b^2}},\quad v:=\frac{-al}{\sqrt{a^2+b^2}},\quad w:=0.$$ (If $a=0,$ you may choose $(u,v,w)=(l,0,0)$.)
Then, the following $n$ points $M_k$ (from $k=0$ to $k=n-1$) are a solution:
$$M_k=P_2+\left(\cos\frac{k2\pi}n\right)V+\left(\sin\frac{k2\pi}n\right)N\times V,$$ where $\times$ is the cross product, i.e. $$N\times V=\left(\beta w-\gamma v,\gamma u-\alpha w,\alpha v-\beta u\right).$$