Suppose that $\mathcal{A}$ is a category with enough projective-injective. The construction of left or right derived functors I know is: for an object $A \in \mathcal{A}$ take a projective (resp. injective) resolution $P^{\bullet}$ of $A$, i.e., $P^{\bullet} \to A$ and apply the functor $F$ and then take homology or cohomology in each degree.
In the special case of the category of $R$-modules, taking $B \in \mathrm{Mod}_R$ we have $\mathrm{Hom}(B,\bullet)$ or $\mathrm{Hom}(\bullet,B)$ from which we can construct $\mathrm{Ext}_R^i(B,\bullet)$ or $\mathrm{Ext}_R^i(\bullet,B)$. I have trouble understanding the following:
If I want to construct $\mathrm{Ext}_R^i(B,A)$, since $\mathrm{Hom}(B,\bullet)$ is left exact, I take a injective resolution of $A$, and then $\mathrm{Hom}(B,\bullet)$ to create $\mathrm{Ext}_R^i(A,B)$.
On the other side the controvariant functor $\mathrm{Hom}(\bullet,B)$ requires me to take a projective resolution of $A$ and do it similarly.
So my question is the following: is there any particular reason to do so? It seems to me that even if $\mathrm{Hom}(\bullet,B)$ is left exact, I could take an injective resolution $Q^{\bullet}$ of $A$, take $\mathrm{Hom}(\bullet,B)$, obtaining $\cdots \to \mathrm{Hom}(Q_n,B) \to \cdots \to \mathrm{Hom}(Q_0,B) \to 0$? Is this not allowed? Is the exactness of the contravariant functor $\mathrm{Hom}(\bullet,B)$ hidden somewhere? Is there a problem on “where to start”?
Any help or clarification would be appreciated.