已知向量a=(1,x-4),向量b=(x^2,3x/2),x属于【-4,2】,求向量a*向量b的最大值
问题描述：
已知向量a=(1,x-4),向量b=(x^2,3x/2),x属于【-4,2】,求向量a*向量b的最大值
并求此时向量a,向量b的夹角的大小
答
a.b
=(1,x-4).(x^2,3x/2)
= x^2+(3x^2-12x)/2
= (5x^2-12x )/2
(a.b)' = (10x-12)/2 =0
x= 6/5
(a.b)'' >0 ( min)
a.b at x= -4
= (5(16)+48 )/2
=64
a.b at x= 2
=(5(4)-12(2) )/2
=-2
max a.b = 64
a.b = 64
at x= -4
a = (1,-8) b=(16,-6)
|a| = √65,|b|= 2√73
a.b = |a||b|cosy
64 =√65(2√73) cosy
y = arccos (32/√4745)