Question

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?

Answer 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