已知向量a=(sinx,1),b=(1,cosx),-π/2

已知向量a=(sinx,1),b=(1,cosx),-π/2
已知向量a=(sinx,1),b=(1,cosx),-π/2

已知向量a=(sinx,1),b=(1,cosx),-π/2
a⊥b
a~b=sinx-cosx=√2 sin(x-π/4)=0
-3π/4x=π/4
|a+b|^2=3+2√2 sin(x-π/4)<=3+2√2
|a+b|的最大值 1+√2

2

a*b=sinx*cosx-1=0
sin2x=1/2
x范围
2x=π/6,5π/6
解得x=π/12,5π/12
|a+b|= √sinx^2+1+cos^2+1+2cosx+2sinx=√3+2(cosx+sinx)
=√3+2√2sin(x+π/4)
-π/4所以当sin(x+π/4)= π/2时有最大值
√3+2√2