设a=(x1,y1),b=(x2,y2),c=(x3,y3)
第一个:a·b=x1x2+y1y2,b·a=x2x1+y2y1,所以,a·b=b·a
第二个:(ka)·b=(kx1,ky1)·(x2,y2)=kx1x2+ky1y2
k(a·b)=k(x1x2+y1y2)=kx1x2+ky1y2
a·(kb)=(x1,y1)·(kx2,ky2)=kx1x2+ky1y2
显然,三者相等
第三个:
(a+b)·c=(x1+x2,y1+y2)·(x3,y3)
=(x1+x2)x3+(y1+y2)y3
=x1x3+x2x3+y1y3+y2y3
=x1x3+y1y3+(x2x3+y2y3)
=a·c+b·c