3. Let vectors $a, b, x, y$ in the plane satisfy the relations $a=x-y, b=2x+y$. Suppose the magnitudes of $a$ and $b$ are both 1, and they are perpendicular to each other. Then the angle between $\boldsymbol{x}$ and $\boldsymbol{y}$ is $\qquad$
3. $\pi-\operatorname{arcos} \frac{\sqrt{10}}{10}$.

Given $x=\frac{a+b}{3}, y=\frac{b-2 a}{3}$.
Let the angle between $\boldsymbol{x}$ and $\boldsymbol{y}$ be $\theta$, then $\cos \theta=\frac{\boldsymbol{x} \cdot \boldsymbol{y}}{|\boldsymbol{x}| \cdot|\boldsymbol{y}|}=-\frac{\sqrt{10}}{10}$.
Therefore, $\theta=\pi-\operatorname{arcos} \frac{\sqrt{10}}{10}$.