I came along this question while trying to simplify the solving of intersection point of lines and surfaces. I am not sure if this will help my initial problem but I am now interested if there is a solution to this. I briefly read about this transformation in a physics paper, but can't figure out how it might be done.
In $\mathbb {R}^3$ there is a line, for example
$L=\left( \begin{array}{c} 1\\ 1\\ 2 \end{array} \right) +v\, \left( \begin{array}{c} 1/\sqrt{6}\\ 1/\sqrt{6}\\ 2/\sqrt{6} \end{array} \right)$
and a "moving" coordinate system along some curve $c(s)$ $X(s)=\left( \begin{array}{c} \cos(s)\\ 0\\ -\sin(s) \end{array} \right)$, $Y(s)=\left( \begin{array}{c} 0\\ 1\\ 0 \end{array} \right)$, $Z(s)=\left( \begin{array}{c} \sin(s)\\ 0\\ \cos(s) \end{array} \right)$
I know the line will be a curve in this coordinate system, but I don't know how to calculate the equation. If there was a Cartesian representation of the line, the transformation would be obvious. But the Cartesian representation of a line consists of two equations or the formula
$\frac{x-1}{1/\sqrt{6}}=\frac{y-1}{1/\sqrt{6}}=\frac{z-2}{2/\sqrt{6}}$
how can one obtain from this one equation for the line in the other coordinate system?
You have an orthonormal basis in the frame $\{X(s), Y(s), Z(s) \}$ i.e all are perpendicular and unit length. Hence, if we let $L(v) = p + tv$ as you've defined then at $L(v)$ we have the vector $v$ and so,
$$ v= (X(s) \cdot v) X(s) + (Y(s) \cdot v) Y(s) + (Z(s) \cdot v) Z(s)$$
which gives the representation of your line in this moving frame. How I got to the above is just by observing that,
$$ v = a X(s) + b Y(s) + c Z(s)$$
and now take dot products with $X(s), Y(s), Z(s)$ using the fact that they are pairwise orthogonal and unit length to recover $a,b,$ and $c$. Just a fun fact, whenever you have such a basis, we call the above expansion, an orthonormal expansion (see "Elementary Differential Geometry" by B. O'Neill).