2. Given that $a, b$ are non-zero vectors and satisfy $(a-2 b) \perp a,(b-2 a) \perp b$, then the angle between $a$ and $b$ is ( )
A. $\frac{\pi}{6}$
B. $\frac{\pi}{3}$
C. $\frac{2 \pi}{3}$
D. $\frac{5 \pi}{6}$
2. B Hint: Since $|a-2 b| \perp a$ then $(a-2 b) \cdot a=0$ so $|a|^{2}=2 a \cdot b$ Similarly $|b|^{2}=2 a$ - $\boldsymbol{b}$ so $|\boldsymbol{a}|=|\boldsymbol{b}|=\sqrt{2 \boldsymbol{a} \cdot \boldsymbol{b}}$ so $\cos \theta=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}||\boldsymbol{b}|}=\frac{1}{2}$ hence $\theta=\frac{\pi}{3}$