2. Given $\boldsymbol{a}=\left(\cos \frac{2}{3} \pi, \sin \frac{2}{3} \pi\right), \overrightarrow{O A}=\boldsymbol{a}-$ $\boldsymbol{b}, \overrightarrow{O B}=\boldsymbol{a}+\boldsymbol{b}$, if $\triangle O A B$ is an isosceles right triangle with $O$ as the right-angle vertex, then the area of $\triangle O A B$ is $\qquad$
2. 1 .

Solution: Let vector $\boldsymbol{b}=(x, y)$, then $\left\{\begin{array}{l}(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{a}-\boldsymbol{b})=0 \\ |\boldsymbol{a}+\boldsymbol{a}|=|\boldsymbol{a}-\boldsymbol{b}|\end{array}\right.$,
which means
$$
\begin{array}{l}
\left\{\begin{array}{l}
\left(x-\frac{1}{2}, y+\frac{\sqrt{3}}{2}\right) \cdot\left(-x-\frac{1}{2},-y+\frac{\sqrt{3}}{2}\right)=0 \\
\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{\sqrt{3}}{2}\right)^{2}=\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{\sqrt{3}}{2}\right)^{2}
\end{array}\right. \\
\text { i.e., }\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
x=\sqrt{3} y
\end{array} \therefore b=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \text { or }\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right),\right. \\
\therefore S_{\triangle M O B}=\frac{1}{2}|a+b||a-b|=1 .
\end{array}
$$