For a $\mathscr{C}^{1}$ curve $c:(-\epsilon,\epsilon) \to \mathbb{R}^{n+1} \setminus \{0\}$, let $\overline{c} = \pi \circ c$, where $\pi: \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{R}P^n$ is the natural projection.
Show that the 1-jets of two curves $\overline{c}_1, \overline{c}_2$ agree $\Leftrightarrow \: \exists \lambda \in \mathbb{R} \setminus \{0\}$, such that $c_2(0) = \lambda \cdot c_1(0)$ and $(\dot{c_2}(0)-\lambda \cdot \dot{c_1}(0) )\in \mathbb{R} \cdot c_2(0)$.
My ideas so far:
$"\Rightarrow"$ $c_1(0) \sim c_2(0)$ in $\mathbb{R}P^n$ $\Leftrightarrow c_1(0) = \lambda \cdot c_2(0)$ for some $\lambda \neq 0$.
And then for the first derivatives: $d \pi_{c_1(0)}\circ c'_1(0) = d\pi_{c_2(0)}\circ c'_2(0)$. But I don't see how this leads to the necessary condition.
Thanks in advance for any help !
As you said, $\bar c_1(0)=\bar c_2(0)=\bar P$ if and only if $P=c_1(0)$ and $Q=c_2(0)$ are nonzero multiples of one another. Now we want to think about $d\pi_P$ and $d\pi_Q$. These both map $\Bbb R^{n+1}$ onto $T_{\bar P}\Bbb RP^n$, with kernel spanned by $P$.
If $Q=\lambda P$, then the curves $c_2$ and $\lambda c_1$ both pass through $Q$ at time $0$ and $d\pi_Q(\dot c_2(0)) = d\pi_Q(\lambda \dot c_1(0))$ if and only if $\dot c_2(0)=\lambda\dot c_1(0)+\mu Q$ for some scalar $\mu$. This is precisely the condition in the statement.