由题意得:a·b=0(a-c)(b-c)=0a·b-a·c-b·c+c^2=0c^2-ac-bc=0|c|^2-|a||c|cosA-|b||c|cos(π/2-A)=0|c|^2-|a||c|cosA-|b||c|sinA=0|c|(|c|-|a|cosA-|b|sinA)=0|c|=0(舍),|c|=|a|cosA+|b|sinA=cosA+sinA=√2sin(A+π/4)因为0π/4√2/2所以1<√2sin(A+π/4) ≤√2即|c|的取值范围是(1, √2].