3. In $\triangle A B C$, $A B=6, A C=8, \angle B A C=$ $90^{\circ}, A D, B E$ are the medians to sides $B C, A C$ respectively. Then the cosine of the angle between vectors $\overrightarrow{A D}$ and $\overrightarrow{B E}$ is equal to ( ).
(A) $\frac{\sqrt{13}}{65}$
(B) $\frac{\sqrt{3}}{2}$
(C) $-\frac{\sqrt{13}}{65}$
(D) 0
3. C.

As shown in Figure 3, with $A$ as the origin and the two perpendicular sides as coordinate axes, we establish a rectangular coordinate system. Thus, $A(0,0)$, $B(6,0)$, $C(0,8)$, $D(3,4)$, and $E(0,4)$. Therefore,
$$
\begin{aligned}
\overrightarrow{A D} & =(3,4), \\
\overrightarrow{B E} & =(-6,4). \\
& \text { Hence } \cos \langle\overrightarrow{A D}, \overrightarrow{B E}\rangle \\
= & \frac{3 \times(-6)+4 \times 4}{\sqrt{3^{2}+4^{2}} \times \sqrt{(-6)^{2}+4^{2}}}=-\frac{\sqrt{13}}{65} .
\end{aligned}
$$