Consider $\Bbb L^3 = (\Bbb R^3, \langle , \rangle)$, with the convention $$\langle (x_1,y_1,z_1), (x_2,y_2,z_2)\rangle = x_1x_2+y_1y_2 - z_1z_2$$ and $\| v \| = \sqrt{|\langle v, v \rangle|}$.
Let $\alpha: I \subset \Bbb R\rightarrow \Bbb L^3$ be a spacelike curve, parametrized by arc-length, and $\mathbf{T}(s) := \alpha '(s)$ be the tangent vector to $\alpha$ at $s$. Suppose $\mathbf{T}'(s)$ is lightlike for all $s$. Then we define the normal vector $\mathbf{N}(s) := \mathbf{T}'(s)$, and the binormal vector $\mathbf{B}(s)$ as the unique lightlike vector orthogonal to $\mathbf{T}(s)$ such that $\langle \mathbf{N}(s), \mathbf{B}(s) \rangle = 1$. We do not define the curvature in this case. I have no problem accepting these definitions, and I can even prove the Frenet-Serret equations: $$\begin{pmatrix} \mathbf{T}' \\ \mathbf{N}' \\ \mathbf{B}' \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & \tau & 0 \\ -1 & 0 & -\tau \end{pmatrix} \begin{pmatrix} \mathbf{T} \\ \mathbf{N} \\ \mathbf{B} \end{pmatrix}$$ where $\tau$ will be the torsion of the curve.
However, I am having some trouble understanding how to compute the binormal vector in practical cases. Surely I could write $\alpha(s) = (x(s), y(s), z(s))$, and $\mathbf{B}(s) = (b_1(s), b_2(s), b_3(s))$, and try to solve the system $$\left\{\begin{array}{c} b_1(s)~x'(s) + b_2(s)~y'(s) - b_3(s)~z'(s) = 0 \\ b_1(s)~x''(s) + b_2(s)~y''(s) - b_3(s)~z''(s) = 1 \\ (b_1(s))^2 + (b_2(s))^2 - (b_3(s))^2 = 0 \end{array} \right.$$
I think that in general, it's very hard to solve this. Can someone give some insight about this, or make an example?
I have the same problem if $\alpha$ is lightlike and pseudo-parametrized by arc-length, but if I can understand this case, I can probably manage the other one by myself. There shouldn't be much difference. Thank you!
(If it happens to help, I'm using this text)