For a vector bundle $V$ on a projective variety $X$, let $\Bbb P(V) $ be the projective bundle of hyperplanes. Call $V$ a very ample if $\mathcal O_{\Bbb P(V)}(1)$ is very ample on $\Bbb P(V)$.
Let $k$ be a field and let $G=G(r,p)$ be the Grassmanian of the $r$-dimentional quotients of $k^p$, with $r,p \in \Bbb Z_{>0}$
Let $E$ be the quotient vector bundle of rank $r$ on $G$.
Is there any reason why $\operatorname{det}E $ would be very ample? How can we see geometrically that the corresponding divisor it linearly equivalent to an hyperplane?