Question

已知向量a=（cos3x/2，sin3x/2），向量b=（cosx/2，-sinx/2），向量c=（根号3，-1）

（1）当a⊥b时，求x的值的集合；
（2）求|a-c|的最大值。

Answer 1

(1)
a⊥b
=> a.b=0
(cos(3x/2), sin(3x/2).(cos(x/2), -sin(x/2))=0
cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)=0
cos2x=0
2x = nπ+π/2
x = nπ/2 + π/4 n=0,1,2,...
x的值的集合 = { x | x= nπ/2 + π/4, n=0,1,2,....}

(2)
a-c = ( cos(3x/2)-cos(x/2), sin(3x/2)+sin(x/2))
|a-c|^2
= (cos(3x/2)-cos(x/2))^2+ (sin(3x/2)+sin(x/2))^2
= 2 -2(cos(3x/2)cos(x/2))- (sin(3x/2)+sin(x/2))
= 2-2cos2x
|a-c|^2的最大值 at cos2x =-1
max |a-c| = 2