已知向量a,b满足|a|=1,|b|=2,向量a与b的夹角为120度,求使a+kb与ka+b的夹角为锐角的实数k的范围?

2025-03-05 01:01:49
推荐回答(1个)
回答1:

k不等于0
(a+kb).(ka+b)
=k|a|^2 + k|b|^2 + (1+k^2)|a||b|cos120度
= k + 2k-(1+k^2)
= -k^2+3k-1 >0
k^2-3k+1<0
(3-√5)/2a+kb与ka+b的夹角为锐角
(3-√5)/2