a+b与2a-b垂直,a-2b与2a+b垂直,所以有
(a+b)(2a-b)=2a^2+ab-b^2=0(1)
(a-2b)(2a+b)=2a^2-3ab-2b^2=0(2)
(1)*3+(2)得
6a^2-3b^2+2a^2-2b^2=0
a^2=(5/8)b^2
代回(1)得
ab=b^2-2a^2=-b^2/4
cos(a,b)
=ab/|a|*|b|
=(-b^2/4)/(根号(5/8)b^2)|b|
=-2/根号40
=-(根号10)/10