If I have a line in 3D space $AB$ where the gradient, $A$ and $B$ are known, how can I find the point at a given angle degrees on the circumference of the circle normal to $AB$ with centre $B$?
Edit: The angle doesn’t really matter, just want to know how to do this in principle.
Edit 2: Here is a figure of what I mean.
Just find two unit vectors $\vec u$ and $\vec v$, perpendicular between them and to line $AB$. You can then write a parametric equation for the circle as: $$ B+r\cos\theta\ \vec u+r\sin\theta\ \vec v. $$ Example.
If $A=(0,0,1)$ and $B=(2,2,0)$ then we can conveniently define a unit vector with direction $AB$ as $$ \vec n={A-B\over AB}={1\over3}(-2,-2,1) $$ and $$ \vec u={1\over\sqrt2}(1,-1,0), \quad \vec v=\vec n\times\vec u={1\over3\sqrt2}(1,1,4). $$