Is the canonical $(n-1)$-plane bundle over the deleted total space of an orientable $n$-plane bundle orientable?

Let $\omega : E \to B$ be a real $n$-plane bundle, and $E_0$ the subspace of $E$ containing all the non-zero vectors. We can construct an $(n-1)$-plane bundle on $E_0$, where the fiber above a vector $v$ above a point $b \in B$ is simply $\omega^{-1}(b)/\mathbb{R}v$. If $\omega$ is orientable, is this vector bundle over $E_0$ orientable?

I think the answer is no, but it seems hard to find an easy counterexample. I'm interested in this, because the complex version of this construction is useful to define the Chern classes, and I wondered if this is one of the reason this construction is not used in the real case with integer coefficients. Another reason is that the Gysin sequence doesn't simplify nicely in dimension $n-1$ of course.

The pullback of $\omega$ via $E_0 \rightarrow B$ is orientable, and the line bundle with fiber over $v$ being $\operatorname{span}(v)$ is trivial since it has an obvious non-zero section. Hence, the quotient bundle you describe is an orientable bundle modulo an orientable subbundle. Since all short exact sequences of vector bundles split, we know that this is equivalent to saying the sum of the quotient bundle and the trivial line bundle is isomorphic to the original bundle.

The direct sum of an orientable bundle with a nonorientable bundle is always nonorientable, so we deduce that the quotient bundle must be orientable.

There are universal constructions that work for both Stiefel-Whitney and Chern classes, but I am not sure if this one can be adapted or not.