3. As shown in Figure 1, the side length of rhombus $A B C D$ is $1, \angle A B C=$ $120^{\circ}$. If $E$ is any point on the extension of $B C$, and $A E$ intersects $C D$ at point $F$, then the angle between vectors $\overrightarrow{B F}$ and $\overrightarrow{E D}$ is $\qquad$
3. $120^{\circ}$.

As shown in Figure 3, establish a Cartesian coordinate system, then $A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), B(0,0), C(1,0), D\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Let $E(a, 0)(a>1)$, then $l_{C D}: y=-\sqrt{3}(x-1), l_{A E}: y=-\frac{\sqrt{3}}{2 a+1}(x-a)$.

By solving the two equations simultaneously, we get the intersection point $F\left(\frac{a+1}{2 a}, \frac{\sqrt{3}(a-1)}{2 a}\right)$ of line $C D$ and line $A E$. Thus, $\overrightarrow{B F}=$ $\left(\frac{a+1}{2 a}, \frac{\sqrt{3}(a-1)}{2 a}\right), \overrightarrow{E D}=\left(\frac{1-2 a}{2}, \frac{\sqrt{3}}{2}\right) \cdot \overrightarrow{B F} \cdot \overrightarrow{E D}=-\frac{a^{2}-a+1}{2 a},|\overrightarrow{B F}|=\frac{\sqrt{a^{2}-a+1}}{a}$ $|\overrightarrow{E D}|=\sqrt{a^{2}-a+1}$
Let the angle between $\overrightarrow{B F}$ and $\overrightarrow{E D}$ be $\theta$, then $\cos \theta=\frac{\overrightarrow{B F} \cdot \overrightarrow{E D}}{|\overrightarrow{B F}| \cdot|\overrightarrow{E D}|}=-\frac{1}{2}$, which means $\theta=120^{\circ}$.