Is the image of a curve contained in a hyperplane if and only if $\det (\gamma'(t),\gamma''(t), \dots, \gamma^{(n)}(t)) = 0$? Let $\gamma : [0,1] \to \mathbb R^n$ be a smooth curve. I'm wondering if the following statement is true:
The image of $\gamma$ is contained in a hyperplane if and only if
$$
\det (\gamma'(t),\gamma''(t), \dots, \gamma^{(n)}(t)) = 0
$$
for all $t \in [0,1]$. I know that the image of $\gamma$ is contained in a hyperplane if and only if the torsion is zero. But is the above equivalent to this statement? Note that we are in $\mathbb R^n$ and $n$ is not necessarily equal to 3.
 A: $\bf{Fact:}$
Let $\phi_1$, $\ldots$, $\phi_m$, $\phi$ $\colon I \to \mathbb{R}$  be smooth functions such that
$$W_m \colon = W(\phi_1, \ldots, \phi_m)(t) = \left| \begin{matrix} \phi_1 &\phi_2 & \ldots & \phi_m \\ 
\phi'_1 &\phi'_2 & \ldots & \phi'_m \\ 
\ldots & \ldots & \ldots & \ldots \\ 
\phi_1^{(m-1)} &\phi^{(m-1)}_2 & \ldots & \phi^{(m-1)}_m
\end{matrix} \right | $$
has no zeroes, but the determinant
$$W_{m+1} \colon =W(\phi_1, \ldots, \phi_m, \phi)(t)= \left| \begin{matrix} \phi_1 &\phi_2 & \ldots & \phi \\ 
\phi'_1 &\phi'_2 & \ldots & \phi' \\ 
\ldots & \ldots & \ldots & \ldots \\ 
\phi_1^{(m)} &\phi^{(m)}_2 & \ldots & \phi^{(m)}
\end{matrix} \right | \equiv 0$$
Then there exist constants $c_1$, $\ldots$, $c_m$ such that
$$\phi(t) = \sum_{i=1}^m c_i \cdot \phi_i(t)$$
$\bf{Proof:}$ Let's write
$$\psi(t) \colon= \left ( \begin{matrix} \phi(t) \\ \phi'(t) \\ \ldots\\ \phi^{(m-1)}(t) \end{matrix} \right)= W_m(t) \cdot c(t) $$
for some $c\colon I \to \mathbb{R}^m$. We'll show that $c$ is a constant function. Indeed, taking the derivative in the above we get
$$\psi'(t) = W_m'(t) \cdot c(t) + W_m(t) \cdot c'(t)$$
But note that since $W_{m+1}$ has dependent columns we have
$$\psi'(t)=W'_m(t)\cdot c(t)$$
We conclude that $W_m(t) \cdot c'(t)\equiv 0$ and so $c'(t)\equiv 0$.
Now apply this result to the matrix of the derivatives of the components. We conclude that the derivatives are linearly dependent, so
$$a_1\cdot  \gamma_1(t) + \cdots + a_n \cdot \gamma_n(t) + b \equiv 0 $$
