Let $G\subseteq \mathcal{O}(\mathbb{R}^3)$ (orthogonal transformations). For a reflection $S\in G$ through a hyperplane $\mathcal{P}\subset\mathbb{R}^3$ we call the two unit vectors $\pm r$ that are perpendicular to $\mathcal P$ roots of $G$. Then we denote $\Delta$ as the set of all roots for a Group and define $\Delta_t^+:=\{r\in\Delta\mid \langle t,r\rangle >0\}$ for some $t\in\mathbb{R}^3$. Furthermore we choose a base $\Pi_t=\{r_1,r_2,r_3\}\subseteq\Delta_t^+$ so that every $r\in\Delta_t^+$ can be written as a non-negative linear combination of elements in $\Pi_t$.
Now let $\Pi_t^*=\{s_1,s_2,s_3\}$ be the dual basis of $\Pi_t$ in $\mathbb{R}^3$ and $F_t=\{\lambda_1 s_1+\lambda _2 s_2+\lambda_3 s_3,\ \lambda_i>0\}$.
Show that $F_t=\{u\in \mathbb{R}^3\mid \Pi_u=\Pi_t\}$.
Proof. Since $\langle s_i\mid r_j\rangle=\delta_{ij}$ we see $\langle u,r_j\rangle=\sum_i\lambda_i \langle s_i\mid r_j\rangle=\lambda_j$ for any $u\in F_t$. So $u=\sum_i\langle u\mid r_i\rangle s_i$. Thus we can write $F_t=\{u\in \mathbb{R}^3\mid \langle u\mid r_i\rangle >0\text{ for all }r_i\in\Pi_t\}=\bigcap\limits_{i=1}^3\{u\in\mathbb{R}^3\mid \langle u\mid r_i\rangle >0\}$. So basically it's left to show that any $u\in\mathbb{R}^3$ that suffices $\langle u\mid r_i\rangle>0$ for $i=1,2,3$ generates the same base for $\Delta_t^+$. Here is where I'm stuck.
*This is an exercise (4.11) in the book "Finite Reflection Groups" by Grove and Benson.