# The relationship between plane curves and the derivative of the Wronskian

I have found a theorem but I did not understand the proof. I'm looking for a clarification of the proof or a different proof.

Let $f_1, f_2, f_3$ be the three components of a curve in $R^3$ parameterized by $t$. Let $W$ be the Wronskian matrix of $f$. whose $(i,j)$ entry is $f_j^{(i-1)}$.

Theorem: If $|W'| = 0$ then $f$ is a planar curve i.e. it satisfies $a_1 f_1 + a_2 f_2 + a_3 f_3 + a_0 = 0$ for some constant $a_i$s.

The proof goes something like this:
We can find $b_i$ which are ordinary functions of $t$ such that $W' b = 0$, that is, $\sum b_i f_i' = 0, \sum b_i f_i'' = 0, \sum b_i f_i''' = 0$.
Although the proof does not explain why such functions $b_i$ exist, I imagine I could watch the null space of $W'$ evolve smoothly with time, and select an evolving vector $b$ living inside that nullspace.
By takng the derivative of the equations, we can arrive at $\sum b_i'f_i' = 0, \sum b_i f_i'' = 0$. Then the proof says something I don't understand: "The $b$'s are proportional respectively to the cofactors of the elements of the last row in the determinant of $W'$. The same is true of the $b'$'s"
Then we get $\frac{b_1'}{b_1} = \frac{b_2'}{b_2} =\frac{b_3'}{b_3} = \phi(t)$ and then $b_i = a_i e^\phi$ so $\sum a_i e^\phi f_i' = 0$ so $e^\phi\sum a_i f_i' = 0$ and then we remove the exponential and integrate to get the final result.

$W'=W(f'), f=(f_1,f_2,f_3)$
If a set of analytic functions has a wronskian of $0$ then the functions are linearly dependent, since the $0^{th}$ to $n-1^{th}$ derivatives of each function can be considered as components of a vector, and checking if the determinant is $0$ is a standard test of linear dependence. If they are linearly dependent, there will be one coefficient for each function, $n$ in total, which is roughly why the derivatives up to $n-1$ contain enough information to determine linear dependence/independence.
$W'=0$ implies that there exist $a_1,a_2,a_3$ such that $a_1f_1'+a_2f_2'+a_3f_3'=0$. Integrating gives the required result.