If $(\vec{a}+3 \vec{b})$ is perpendicular to $(7 \vec{a}-5 \vec{b})$ and $(\vec{a}-4 \vec{b})$ is perpendicular to $(7 \vec{a}-2 \vec{b})$, then the angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ (in degrees) is_______.
$(\vec{a}+3 \vec{b}) \perp(7 \vec{a}-5 \vec{b})$
$(\vec{a}+3 \vec{b}) \cdot(7 \vec{a}-5 \vec{b})=0$
$7|\vec{a}|^{2}-15|\vec{b}|^{2}+16 \vec{a} \cdot \vec{b}=0$............(1)
$(\vec{a}-4 \vec{b}) \cdot(7 \vec{a}-2 \vec{b})=0$
$7|\vec{a}|^{2}+8|\vec{b}|^{2}-30 \vec{a} \cdot \vec{b}=0$.............(2)
from (1) & (2)
$|\vec{a}|=|\vec{b}|$
$\cos \theta=\frac{|\vec{b}|}{2|\vec{a}|} \quad \therefore \theta=60^{\circ}$
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