1. Let $A$ and $B$ be two moving points on the ellipse $\frac{x^{2}}{2}+y^{2}=1$, and $O$ be the origin. Also, $\overrightarrow{O A} \cdot \overrightarrow{O B}=0$. Let point $P$ be on $AB$, and $O P \perp A B$. Find the value of $|O P|$.
Three, 1. Let $A(a \cos \alpha, a \sin \alpha), B(-b \sin \alpha, b \cos \alpha)$.
Substituting into the ellipse equation, we get
$$
\begin{array}{l}
a^{2}\left(\frac{1}{2} \cos ^{2} \alpha+\sin ^{2} \alpha\right)=1, b^{2}\left(\frac{1}{2} \sin ^{2} \alpha+\cos ^{2} \alpha\right)=1 \\
\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{2}+1=\frac{3}{2} \Rightarrow \frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{3}{2} .
\end{array}
$$

In the right triangle $\triangle O A B$,
$$
|O P||A B|=|O A||O B| \text {. }
$$

Thus, $|O P|^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}=\frac{2}{3}$.
Therefore, $|O P|=\frac{\sqrt{6}}{3}$.