Consider the Fano threefold $V=V_5$ of index $2$ and degree $5$. Note that $V = \mathrm{Gr}(2,5) \cap \mathbb{P}^6 \subset \mathbb{P}^9$. Let $U$ and $Q$ be the tautological and tautological quotient bundles on $V$ (obtained by restricting the tautological bundles on the Grassmannian to $V$ - I will use the notation $U_G$ and $Q_G$ for these). I have seen that $\mathrm{Hom}(\mathcal{O}_V, U^\vee \otimes Q^\vee) = k^3$, and I'm trying to prove this.
I assume one would do this by taking some kind of a Koszul resolution of $V$ inside $\mathrm{Gr}(2,5)$, and then use the Borel-Bott-Weil theorem. My issue is with the first step (the Koszul resolution/complex).
I know that if we have a rank $r$ vector bundle $E$ on $X$ and a codimension $r$ irreducible subvariety $Z \subset X$ which is the zero locus of some $s \in H^0(E)$, then we have the following Koszul complex: $$ 0 \to \det E^\vee \to \Lambda^{r-1} E^\vee \to \cdots \to E^\vee \to \mathcal{O}_X \to \mathcal{O}_Z \to 0 . $$
However, I'm not sure how to apply this to the case above, since I don't know of a vector bundle on $\mathrm{Gr}(2,5)$ for which the zero locus of a global section is $V$. Once one knows this, I assume that we would then use the Koszul complex above, and tensor by $U_G^\vee \otimes Q_G^\vee$. Then the last term would become $U_G^\vee \otimes Q_G^\vee|_V \cong U^\vee \otimes Q^\vee$. Then we would take the long exact sequence in cohomology and compute $\mathrm{Hom}(\mathcal{O}_V, U^\vee \otimes Q^\vee)$ as required.
Thank you.
Edit: Using Tabes' suggestion in the comments of using the bundle $\mathcal{O}_{\mathbb{P}^9}^{\oplus 3}|_{G}$ (using $G$ for the Grassmannian now), I think the Koszul complex becomes $$ 0 \to \mathcal{O}_G(-3H) \to \mathcal{O}_G(-2H)^{\oplus 3} \to \mathcal{O}_G(-H)^{\oplus 3} \to \mathcal{O}_G \to \mathcal{O}_V \to 0$$ where $H$ is the pullback of the amble divisor on $\mathbb{P}^9$ to $G$. Now tensor the above by $U_G^\vee \otimes Q^\vee_G$. We get $$ 0 \to U^\vee_G(-3H) \otimes Q^\vee \to (U_G^\vee(-2H) \otimes Q_G^\vee)^{\oplus 3} \to (U_G^\vee(-H) \otimes Q_G^\vee)^{\oplus 3} \to U_G^\vee \otimes Q_G^\vee \to $$ $$ \to U^\vee \otimes Q^\vee \to 0 .$$
I think that Borel-Bott-Weil tells us that $H^i(G, U_G^\vee \otimes Q_G^\vee) = 0$ for all $i$, but so far I'm unsure as to what the cohomologies of the twists are. Here, I think I need to know how to write e.g. $U_G^\vee(-H)$ as $\Sigma^\alpha U_G^\vee$ (since every $\mathrm{GL}(V)$-equivariant representation of $\mathrm{Gr}(2, V)$ can be written as $\Sigma^\alpha U_G^\vee \otimes \Sigma^\beta Q^\vee_G$, where in this case $\alpha \in \mathbb{Z}^2$ and $\beta \in \mathbb{Z}^3$). For example, $\Sigma^{(1,0)}U_G^\vee = U_G^\vee$.
I think for a rank $n$ vector bundle $E$ on a variety $X$ the notation $\Sigma^\alpha E$ means taking the principal $\mathrm{GL}(n)$ bundle on $X$, and then $\Sigma^\alpha$ is the representation of weight $\alpha$ of the principal bundle.
I'd be grateful for any help/comments on the above edit. Thanks!