1. As shown in Figure $1, O, A, B$ are three points on a plane, vectors $\boldsymbol{O A}$ $=a, O B=b$. On the plane $A O B$, $P$ is any point on the perpendicular bisector of line segment $A B$, vector $O P=p$, and $|a|=3$, $|b|=2$. Then $p \cdot(a-b)=$
ii. $1 \cdot \frac{5}{2}$.
Let the perpendicular bisector of $AB$ intersect $AB$ at $C$, and connect $OC$. Then $OC=\frac{1}{2}(a+b)$. Therefore,
$$
PC=OC-OP=\frac{1}{2}(a+b)-p.
$$

Since $AB \perp PC$, i.e., $AB \cdot PC=0$, we have,
$(b-a) \cdot\left[\frac{1}{2}(a+b)-p\right]=0$.
Thus, $p \cdot(a-b)=\frac{1}{2}\left(a^{2}-b^{2}\right)=\frac{1}{2}\left(|a|^{2}-|b|^{2}\right)=\frac{5}{2}$.