I have the center of a circle $\vec{c}$ in 3 space and the radius $r$. I also have a unit vector $\hat{v}$ defining the orientation of the plane of the circle.
I wish to parameterize this circle and place a single point on the circle.
My logic is, imagine I have a point in polar coordinates.
I could describe it as
$$\vec{p} = r \cos(\lambda) \hat{i}+ r \sin(\lambda)\hat{j}+0\hat{k}$$
Now the way I'm imagining this is that this point can sweep out the entire circle just by modifying $\lambda$.
Intuitively, I can move this point so that the center of the circle it sweeps out lays at the point of the circle I want.
Thus, I can do
$$\vec{p} = \vec{c} + r \cos(\lambda) \hat{i}+ r \sin(\lambda)\hat{j}+0\hat{k}$$
However this still doesn't give me the right orientation. I wish the plane of the circle to be perpendicular to $\hat{v}$. I am stuck with my initial orientation of my circle and now I wish to reorient the circle along the unit vector $\hat{v}$.
How do I do this?
Method 1: 2 parameters, and you'd get a disc instead of a circle
Take all vectors on the unit ball with center at origin. Take their cross product with $\hat{v}$ (this will leave behind only that part of each vector which is perpendicular to $\hat{v}$. Finally multiply by $r$ and add $\vec{c}$ $$r(\cos(\theta)\cos(\phi)\hat{i}+\cos(\theta)\sin(\phi)\hat{j}+\sin(\theta)\hat{j})\times\hat{v}+\vec{c}$$
Method 2: 1 parameter, and you'd get a circle. But need to choose something first...
Choose any plane (passing through origin) of which $\hat{v}$ is not a part of. Since $\hat{v}$ can't simultaneously be on the XY, YZ and ZX planes, so we assume (WLOG) that XY is our plane of choice. Take the unit circle centered at the origin on that plane. Cross with $\hat{v}$. Normalize to unit magnitude. Multiply by $r$ and add $\vec{c}$. $$r\frac{(\cos(\theta)\hat{i}+\sin(\theta)\hat{j})\times \hat{v}}{|(\cos(\theta)\hat{i}+\sin(\theta)\hat{j})\times \hat{v}|}+\vec{c}$$
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[Please comment if any errors are found.]