Question

已知非零向量ab满足|a-b|=|a+b|=c|b| 则向量a-b'与a+b的夹角最大值是

Answer 1

|a+b|=|a-b|的话,说明：a·b=0

即a⊥b,故：=π/2

而：|a+b|^2=c^2|b|^2,即：|a|^2=(c^2-1)|b|^2

(a+b)·(a-b)=|a|^-|b|^2=(c^2-2)|b|^2

|a-b|^2=|a+b|^2=|a|^2+|b|^2=c^2|b|^2

故：cos=(a+b)·(a-b)/(|a+b|*|a-b|)

=(c^2-2)|b|^2/(c^2|b|^2)=(c^2-2)/c^2|=1-2/c^2

c≥2,故：c^2≥4,即：0