(1)|a|=1,|b|=√3 a+b=(√3,1)(a+b)²=a²+2ab+b²=1+2ab+3=4所以ab=0|a-b|²=a²-2ab+b²=1+3=4所以|a-b|=2(2)(a+b)(a-b)=a²-b²=1-3=-2|a+b|=2 |a-b|=2所以cosθ=[(a+b)(a-b)]/[...