Path on Unit Sphere for which Binormal is Parallel to Position I am wondering if it is possible to produce a path on the unit sphere for which the Binormal vector ${B}$ is parallel to the position vector at some instant (or all instants!).
I tend to think of the binormal as the axis around which a path is instantaneously rotating. (Is that correct?) I was considering curves on the unit sphere, and my intuition said that they would always be rotating about the origin. But the Darboux vector seems to suggest otherwise. It is $$\omega = \tau T + \kappa B$$ and it claims to be the axis around which a path is rotating. What confused me was that it has a $T$ component. The connection is that $T$ is a path on the unit sphere.
Clearly something in my understanding is wrong. I think an answer to the question above would help.

Edit: Let me restate the question now that I have thought about it more. Is it case that the center of curvature for a path on the unit sphere must be the origin, or need not this be true? Under what general conditions is it true?
 A: The binormal of a constant path on the unit sphere is everywhere parallel to the position vector.
The binormal of a regular path $\gamma$ (non-vanishing velocity) on the unit sphere is proportional to the position $\gamma(t)$ if and only if $\gamma''(t)$ is tangent to the sphere. This never happens: In suitable coordinates, $\gamma$ is a reparametrization of
$$
\gamma(t) = \bigl(t, f(t), \sqrt{1 - t^{2} - f(t)^{2}}\bigr)
$$
for some (sufficiently smooth) function $f$ with $f(0) = 0$. A short calculation shows
$$
\frac{d^{2}}{dt^{2}}\bigg|_{t = 0} \sqrt{1 - t^{2} - f(t)^{2}} = -\bigl(1 + f'(0)^{2}\bigr) < 0.
$$
That is, $\gamma''(0)$ has a non-zero third component, hence is not tangent to the sphere at $\gamma(0) = (0, 0, 1)$.

To address the edit, if $\gamma$ is a constant-speed non-great circle $C$ on the unit sphere, then the center of curvature of $\gamma$ is the center of $C$, not the center of the sphere.
The center of curvature of $\gamma$ is the center of the sphere if and only if the normal vector of $\gamma$ is $-\gamma$. When $\gamma$ is a unit-speed path, this is equivalent to saying the geodesic curvature at $\gamma(t)$ is zero.

