|向量a|=2,|向量b|=1，两者夹角为:派/3，求向量2a+3b与3a-b的夹角

由于a、b夹角为 π/3，可以不失一般性的假设
a = 2[cos(π/3) + i*sin(π/3)] = 1 + √3*i
b = 1

A = 2a + 3b
= (2*1 + 3*1) + 2√3*i
= 5 + 2√3*i
斜率：K1 = 2√3/5

B = 3a - b
= (3 - 1) + √3*i
= 2 + √3*i
斜率：K2 = √3/2

tg(B-A)
= (K2 - K1)/(1 + K1*K2)
= (√3/2 - 5√3/2)/(1 - 6/10)
= √3/(2/5)
= 5/2*√3

B-A = arctg(5/2*√3)