2. Given vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $|\boldsymbol{a}|=1,|\boldsymbol{b}|=\sqrt{3}$, and $(3 a-2 b) \perp a$. Then the angle between $a 、 b$ is $\qquad$
2. $\frac{\pi}{6}$.

Let the required angle be $\theta$. Then
$$
0=(3 a-2 b) \cdot a=3|a|^{2}-2|a||b| \cos \theta \text {. }
$$

Combining $|a|=1,|b|=\sqrt{3}$, we have
$$
\cos \theta=\frac{3}{2} \cdot \frac{|a|}{|b|}=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} .
$$