8 Let two non-zero vectors $\vec{a}$ and $\vec{b}$ in the plane be perpendicular to each other, and $|\vec{b}|=1$, then the sum of all real numbers $m$ that make the vectors $\vec{a}+m \vec{b}$ and $\vec{a}+(1-m) \vec{b}$ perpendicular to each other is
81 Hint: Since
$$
\begin{aligned}
0 & =(\vec{a}+m \vec{b}) \cdot[\vec{a}+(1-m) \vec{b}] \\
& =\vec{a}^{2}+\vec{a} \cdot \vec{b}+m(1-m) \vec{b}^{2} \\
& =|\vec{a}|^{2}+m(1-m),
\end{aligned}
$$

i.e., $m^{2}-m-|\vec{a}|^{2}=0$, so by the relationship between roots and coefficients, the sum of all real numbers $m$ that satisfy the condition is 1.