Let $X$ be a complex variety and $\mathscr E$ a locally free sheaf on $X$. Consider the fiber bundles $$ \mathbb P(\mathscr E) \overset{def}= \mathrm{Proj}(\mathrm{Sym}(\mathscr E)), \quad \mathbb A(\mathscr E) = \mathrm{Spec}(\mathrm{Sym}(\mathscr E)). $$ There is a canonical rational map $f : \mathbb A(\mathscr E) \dashrightarrow \mathbb P(\mathscr E)$ sending which can be seen to be the projection $\mathbb A^n \backslash \{0\} \to \mathbb P^{n-1}$ on each fiber. Let $U(\mathscr E) \subseteq \mathbb A(\mathscr E)$ be the maximal open subset on which $f$ is defined as a map (e.g. not a rational map) ; in terms of varieties, $U(\mathscr E)$ is the complement of the image of the zero sections of $\mathbb A(\mathscr E) \to X$.
My question : Does the map $U(\mathscr E) \to \mathbb P(\mathscr E)$ admit a global section? In other words, is there an algebraic way to associate to every line in $\mathscr E$ a vector in that line?
I have a feeling this has to something to do with orientability, but I am not quite sure. The "real" picture I have in mind is $X$ being a point and $\mathscr E$ equal to $\mathbb R^3$, where this section is a half-sphere centered at the origin. However, even there I encounter issues because "half the sphere" is not an algebraic variety. I'm not convinced this result is true, but if it is, it would help me construct some maps over such vector/projective bundles.