3. Let point $Q$ be in the plane $\alpha$ of $\triangle A B C$, and point $P$ be outside the plane $\alpha$. If for any real numbers $x$ and $y, |\overrightarrow{A P}-x \overrightarrow{A B}-y \overrightarrow{A C}| \geq|\overrightarrow{P Q}|$, then the angle $\theta$ between vector $\overrightarrow{P Q}$ and $\overrightarrow{B C}$ is $\qquad$.
Answer $\frac{\pi}{2}$.
Analysis Let $\overrightarrow{A M}=x \overrightarrow{A B}+y \overrightarrow{A C}$, then point $M$ is also in plane $\alpha$, $|\overrightarrow{M P}|=|\overrightarrow{A P}-\overrightarrow{A M}| \geq|\overrightarrow{P Q}|$, so $P Q \perp$ plane $\alpha$, thus the angle between vector $\overrightarrow{P Q}$ and $\overrightarrow{B C}$ is $\frac{\pi}{2}$.