13. Let $s, t$ be two non-zero real numbers, $\vec{a}, \vec{b}$ be unit vectors, and $s \vec{a}+t \vec{b}$ has the same magnitude as $t \vec{a}-s \vec{b}$. Find the angle between vectors $\vec{a}$ and $\vec{b}$.
13. $|s \vec{a}+t \vec{b}|^{2}=|t \vec{a}-s \vec{b}|^{2}$, i.e., $s^{2} \vec{a}^{2}+2 s \vec{a} \cdot \vec{b}+t^{2} \vec{b}^{2}=t^{2} \vec{a}^{2}-2 s t \vec{a} \cdot \vec{b}+s^{2} \vec{b}^{2}$. Given $|\vec{a}|=|\vec{b}|=1$, we get $\vec{a} \cdot \vec{b}=0$, which means the angle between vector $\vec{a}$ and vector $\vec{b}$ is 90.