Question

已知向量a=(cos(-θ),sin(-θ)),b=(cos(π/2-θ),sin(π/2-θ) (1)求证：a⊥b (

Answer 1

1.a=(cos(θ),-sin(θ)),b=(sin(θ),cos(θ)),a*b=0,故a⊥b

2.x*y=k*a^2+t*(t^2+3)b^2=k+t*(t^2+3)=0

（k+t^2)/t=-t^2-3+t=-(t-1/2)^2-11/4

因此最大值是-11/4 没有最小值