I was reading a paper about Bishop frame and try to calculate the Bishop frame for a cubic bezier curve. My plan to do so is to generate an arbitrary adapted frame $\left\{T, N_1, N_2 \right\}$ and then solve the ODE regarding the angle $\theta$ between the Bishop frame and the adapted frame. My problem is, how can I derive the explicit formula for the adapted frame? Note that the adapted frame must be at least 1 order differentiable.
1 proposed the method of applying Gram-Schmidt process to $T$ "and two parallel field", which seems not working because GS process requires independent vectors as input.
The problem can actually be stated in a more general way. Given one vector $T$, how can one generate an orthogonal base containing this vector in an explicit way? In $\mathbb{R}^3$, you can define an "up" vector $U$, normalize $F = U \times T$ to get the "front" vector, then $T \times F$ will be the "right" vector. However, this method has flaws: it will degenerate when $T$ happens to be parallel to $U$. As a result, you have to discuss this situation, making the formula not explicit. What I want is a formula that satisfy $$f(T) \perp T $$ for every possible $T$.
Is that possible?
Also, if that is not possible, I would like to know if there is another way to deal with my original problem.
A partial answer to the problem:
According to this post. The general problem of assigning an orthogonal base containing vector $v$ that is also continuously depend on $v$ is not possible in three dimensional space.
Suppose such function exists (and not trivial), then we have a non vanishing tangent vector field of $S^2$, which contradicts the Hairy ball theorem
A workaround can be found in the comments of the problem. Basicly, we don't need a general formula for every vector since the tangent mapping of the curve corresponds to only a small fraction of $S^2$, we can easily pick a vector that does not lie on the tangent mapping.