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

|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≥