Vector Cross product - Rearranging issue Given Data in question


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*I have following relations in vector space$\begin{eqnarray}n_0^{'}(s)=-\kappa(s) \times n_0(s)\\n_1^{'}(s)=-\kappa(s) \times n_1(s)\\n_2^{'}(s)=-\kappa(s) \times n_2(s)\end{eqnarray}\tag 1$

*$n_0,n_1,n_2$ are orthogonal unit vectors

*$\kappa(s)$ is not a unit vector

*Derivative of a unit vector is a vector that is normal to the given vector. We can normalize it to get normalized normal unit vector. Click here  for Reference if needed for this data
Question


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*Can we rewrite the equation (1) in the following form$\begin{eqnarray}n_0(s)=-\kappa(s) \times \psi_0(s)\\n_1(s)=-\kappa(s) \times \psi_1(s)\\n_2(s)=-\kappa(s) \times \psi_2(s)\end{eqnarray}\tag 2$?
If so what could be the vectors $\psi_0(s),\psi_1(s),\psi_2(s)$ ?
or can we rewrite equation (1) with a LHS in equation (2) irrespective of what is on R.H.S(just avoiding R.H.S condition for more relaxation in complexity of the problem )? 
 A: Alright, let's look for an inverse here (it sounds like one does not exist)--which means your problem is either impossible or requires the constraints that the $\vec{n}$'s are orthogonal unit vectors.
The cross product is restricted to $\mathbb{R}^3$ (assuming reals--but that isn't really a constraint here, I don't think).  So we have:
$$
\langle x_1, y_1, z_1\rangle\times\langle x_2, y_2, z_2\rangle = \langle y_1z_2 - z_1y_2, z_1x_2 - x_1z_2, x_1y_2 - y_1x_2\rangle
$$
Then we have:
\begin{align}
\langle x_3, y_3, z_3\rangle\times\left(\langle x_1, y_1, z_1\rangle\times\langle x_2, y_2, z_2\rangle\right) = & \langle y_3(x_1y_2 - y_1x_2) - z_3(z_1x_2 - x_1z_2),\\
&z_3(y_1z_2 - z_1y_2) - x_3(x_1y_2 - y_1x_2),\\
& x_3(z_1x_2 - x_1z_2) - y_3(y_1z_2 - z_1y_2)\rangle
\end{align}
Now this looks very complicated (and it is--if you need to solve it "as is").  Instead we can imagine that we already know $\vec{n}' = \vec{v}_1\times\vec{v_2}$.  Then this becomes:
\begin{align}
\langle x_3, y_3, z_3\rangle\times\vec{n}' = & \langle y_3n'_z - z_3n'_y, z_3n'_x - x_3n'_z, x_3n'_y - y_3n'_x\rangle
\end{align}
Now, setting this equal to the second argument of the original cross product, we have a set of linear equations for three unknowns ($x_3, y_3, z_3$):
$$
x_2 = y_3n'_z - z_3n'_y \\
y_2 = z_3n'_x - x_3n'_z \\
z_2 = x_3n'_y - y_3n'_x
$$
The matrix gives:
$$
\begin{pmatrix} 0 & n_z' & -n_y' \\
-n_z' & 0 &n_x' \\
n_y' & -n_x' & 0
\end{pmatrix} \times \begin{pmatrix}x_3 \\
y_3 \\
z_3\end{pmatrix} = \begin{pmatrix}x_2 \\
y_2 \\
z_2\end{pmatrix}
$$
There is no solution when the determinate of this matrix is zero (which could happen and if this occurs with $\langle x_2, y_2, z_2\rangle = \vec{0}$ then there are an infinite amount of solutions)--but otherwise you should be able to find an inverse.
Alright, so I just noticed that the determinate here (I'm assuming of all anti-symmetric matices) is zero: $n_z'x_x'n_y'- n_y'n_z'n_x' = 0$.  This means that unless $\langle x_2, y_2, z_2\rangle$ is $\vec{0}$ there is no solution (since a matrix that has a determinant equal to zero only has a solution when the RHS is zero--in which case it has infinite solutions).
I'm going to leave this answer here, although I now realize it does not address your problem at all, rather is just a proof of the previous link that the inverse of the cross product does not exist.
