∵(向量a+向量b)⊥向量b,∴(a+b).b=0.
即,ab+b^2=0.
|a||b|cos+|b|^2=0.
∵|a|=1,∴ |b|^2+|b|cos=0,
|b|(|b|+cos)=0.
又|b|≠0.∴|b|+cos=0.
|b|=-cos.[π/2＜≤π]
當(dāng)向量a與向量b反向共線時(shí),=π,此時(shí)|b|=1.