Let me just set the scene before asking my question.
Let V be a real vector space with basis $u_1,...,u_n $. Let $W$ be a Coxeter group with generating set $S=\{s_1, ..., s_n \}$ and let $m_{ij}=|s_is_j|$. Let $B$ be the bilinear form on $V$ defined by $B(u_i, u_j)=-\cos(\pi / m_{ij})$. Let $\sigma_i$ be the linear map on $V$ where $\sigma_i(x)=x-2B(u_i, x)u_i$.
Then $\sigma_i$ is the reflection with respect to the hyperplane $H_i=\{v \in V : B(u_i, v)=0 \}. $
I am trying to show that the order of $\sigma_i \sigma_j$ is $m_{ij}$.
I am reading a proof of this and it shows if we let $P$ be the span of $u_i$ and $u_j$ then $V=P \oplus P^{\perp}$ where $P^{\perp} = \{u \in V : B(u, v)=0 \text{ for all } v \in P \}.$ Then the quadratic form $Q$ where $Q(v)=B(v, v)$ is positive-definite and so $B$ is nonsingular on $P$.
Since $\sigma_i$ and $\sigma_j$ act trivially on $P^{\perp}$ the order of $\sigma_i \sigma_j$ is just equal to the order of $\sigma_i \sigma_j $ restricted to $P$.
My problem
It then says that since $Q$ is positive definite then $\sigma_i$ and $\sigma_j$ correspond to reflections in the Euclidean plane.
Since $$B(u_i, u_j )=-\cos(\pi /m_{ij})= \cos(\pi- (\pi/m_{ij}))$$ the angle between the unit vectors $u_i$ and $u_j$ is $\pi- \pi/m_{ij}$. I don't understand this bit at all. I don't know why $Q$ being positive-definite means that $\sigma_i$ and $\sigma_j$ correspond to reflections in the Euclidean plane and then I don't get how they then deduce that the angle is $2\pi / m_{ij}$. They then go on to say that then this means that $\sigma_i \sigma_j$ corresponds to a rotation through $2\pi / m_{ij}$, I think this comes just from the fact that composing reflections gives a rotation.
Any help with this?