a1*a2=0,设a3=(x,y,z),由a1,a3正交得x+y+z=0,同理x-2y+z=0,解得y=0,z=-x.∴a3=x*(1,0,-1),x是不等于0的实数。a3要满足齐次线性方程Ax=0,而且经过A初等行变换得基础解系,是因为a1,a2,a3两两正交.
