I have a circle in 3-space (the coordinates of its center, its radius, and a vector normal to its plane). I also have several points on that circle. Now I want to find the unit tangents to the circle at these points directed clockwise with respect to the normal vector.
For this, I need to find every perpendicular vector to my points lying on the circle and eliminate all but one, which is perpendicular in a clockwise manner to the vector that goes through circle center and circumference point.
Can anyone explain how I can do this cause I have no idea how to do it.
Let $C$ be the circle, ${\bf c}\in {\bf R}^3$ be its center, and $\bf n$ be a normal to the plane containing $C$. For any point $\bf p$ on $C$, the displacement vector from $\bf c$ to $\bf p$ is simply ${\bf d}={\bf p}-{\bf c}$, and so the (unit) outward-pointing vector perpendicular to $C$ at $\bf p$ is ${\bf o} = \frac{1}{||{\bf d|}|} {\bf d}$.
Thus the clockwise-pointing tangent to $C$ at $\bf p$ is then $$\begin{eqnarray} {\bf t}&=&\frac{1}{||{\bf n}||}{\bf n}\times{\bf o} \\ &=& \frac{1}{||{\bf n}|| \; ||{\bf p}-{\bf c}||}{\bf n}\times({\bf p}-{\bf c}), \end{eqnarray}$$ where clockwise is defined in terms of observing $C$ along $\bf n$.