a=(3,—4),a+b=(4,—3)
所以b=(4-3,-3-(-4))=(1,1)
cos(a,b)=a·b/(|a||b|)
a·b=3*1+(-4)*1=-1
a的模|a|=√(3^2+(-4)^)=5
b的模|b|=√(1^2+(1)^)=√2
cos(a,b)=a·b/(|a||b|)=-1/5√2=-√2/10
a=(3,—4),a+b=(4,—3)
所以b=(4-3,-3-(-4))=(1,1)
cos(a,b)=a·b/(|a||b|)
a·b=3*1+(-4)*1=-1
a的模|a|=√(3^2+(-4)^)=5
b的模|b|=√(1^2+(1)^)=√2
cos(a,b)=a·b/(|a||b|)=-1/5√2=-√2/10