10 If the vector $\vec{a}+3 \vec{b}$ is perpendicular to the vector $7 \vec{a}-5 \vec{b}$, and the vector $\vec{a}-4 \vec{b}$ is perpendicular to the vector $7 \vec{a}-2 \vec{b}$, then the angle between vector $\vec{a}$ and $\vec{b}$ is $\qquad$.
(10) $\frac{\pi}{3}$ Hint: According to the problem, we have $\left\{\begin{array}{l}(\vec{a}+3 \vec{b}) \cdot(7 \vec{a}-5 \vec{b})=0, \\ (\vec{a}-4 \vec{b}) \cdot(7 \vec{a}-2 \vec{b})=0 .\end{array}\right.$ Therefore, $|\vec{a}|^{2}=$ $|\vec{b}|^{2}=2 \vec{a} \cdot \vec{b}$, so $\cos \langle\vec{a}, \vec{b}\rangle=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{2},\langle\vec{a}, \vec{b}\rangle=\frac{\pi}{3}$.