第1个回答 2013-01-13
设向量a=(x1, y1) , b=(x2, y2)
(1)a⊥b<=>a.b=lallblcos<a,b>=0
推理过程:a=x1i+y1j,b=x2i+y2j,(i,j为单位向量,ij=0),
a.b=(x1i+y1j).(x2i+y2j)=x1x2lil^2+[x1y2+x2y1]ij+y1y2ljl^2=x1x2+y1y2
所以:x1x2+y1y2=0
(2)a//b<=>a=λb ,(b≠0)
即(x1,y1)=λ(x2,y2)=(λx2, λy2),
所以x1= λx2,y1=λy2=>λ=x1/x2=y1/y2
=> x1y2-x2y1=0