10. $\mathrm{U}$ knows
$$
a+3 b \perp 7 a-5 b \text {, and } a-4 b \perp 7 a-2 \text {. }
$$

Then the angle between $a$ and $b$ is $\qquad$ .
10. $\frac{\pi}{3}$.

According to the problem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
(a+3 b) \cdot(7 a-5 b)=0, \\
(a-4 b) \cdot(7 a-2 b)=0
\end{array}\right. \\
\Rightarrow|a|^{2}=|b|^{2}=2 a \cdot b \\
\Rightarrow \cos \langle a, b\rangle=\frac{a \cdot b}{|a||b|}=\frac{1}{2} \\
\Rightarrow\langle a, b\rangle=\frac{\pi}{3} . \\
\end{array}
$$