(a+b)·(2a-b)=2a^2+ a·b - b^2=0
(a-2b)·(2a+b)=2a^2 - 3a·b - 2b^2=0
两式相减,得：
b^2 = -4a·b,
∴a^2 = (-5/2)a·b,
∴a·b=|a|·|b|·cos ≤ 0,
即cos ≤0
∴cos = (a·b)/(|a|·|b|)
两边平方,得：
cos^2
= (a·b)^2/(a^2·b^2)
=(a·b)^2/[(-4a·b)(-5a/2 · b)]
=1/10
∴cos= - √10/10
=arccos(-√10/10)