x=(√3+(2t-3)/2,-1+√3/2(2t-3)),y=(-k√3+t/2,k+√3/2t)
x*y=[√3+(2t-3)/2][-k√3+t/2]+[-1+√3/2(2t-3)][k+√3/2t]
=2t^2-21t/4-(4+3√3/2)k
=0
所以k,t的关系是：
k=(2t^2-21t/4)/(4+3√3/2)
(-21/4)^2+4*2*(4+3√3/2)k>=0
k>=-441/(128+48√3)