Question
Let $\gamma$ be the tautological line bundle of the $n$-dimensional real projective space $\mathbb{P}^n$. Is $\gamma$ is orientable as a vector bundle?
My guess
I only consider in case of $n=1$.
Let $n=1$, and consider the one dimensional projective space. let $\nu$ be a normal bundle on the two dimensional sphere $\mathbb{S}^1 \subset \mathbb{R}^2$. Then, there is a bundle map $ \nu \to \gamma$. Then the pull back of the cohomology class $u_F$ of $\gamma$ also gives a orientation of the bundle $\nu$. However, the antipodal points of $\mathbb{S}^1$ are sended to the same point of $\mathbb{P}^1$ and thus, such orientation will be given by some same vector in $\mathbb{E}^2$. But it is impossible, because the orientation of antipodal points should be opposite directions. Thus $\gamma$ is not orientable as a vector bundle.
Definition of orientation 1 Each fiber $F$ of $\gamma$ is given an orientation as an vector space and ,in addition, for each $p \in \mathbb{P}^n$ there are a neighborhood $N$ of $\mathbb{P}^n$ centered at $p$ and a local section $s$ of $\gamma$ defined on $N$ such that the value of section $s$ at each $q \in N$ gives the orientation of the fiber at $q$.
Definition of orientation 2 For each fiber, there is a generator $u_F \in H^1(F,F-0,\mathbb{Z})$ such that for each $q \in \mathbb{P}^n$ there is an element $u \in H^1(\pi^{-1}(N),\pi^{-1}(N)-0,\mathbb{Z})$ whose restriction to $F$ is coincides with $u_F$.
