Question

已知a,b不共線,證明c=x₁a+y₁b與d=x₂a+y₂b為共線向量的充要條件為x₁y₂-x₂y₁=0.

Answer 1

證明:設(shè)a=(a₁,b₁),b=(a₂,b₂),

∵a,b不共線,∴a₁b₂-a₂b₁≠0.

又c=x₁(a₁,b₁)+y₁(a₂,b₂)=(a₁x₁+a₂y₁,b₁x₁+b₂y₁),

d=x₂(a₁,b₁)+y₂(a₂,b₂)=(a₁x₂+a₂y₂,b₁x₂+b₂y₂),

c∥d(a₁x₁+a₂y₁)(b₁x₂+b₂y₂)-(a₁x₂+a₂y₂)(b₁x₁+b₂y₁)=0(a₁b₂-a₂b₁)(x₁y₂-y₁x₂)=0.

∵a₁b₂-a₂b₁≠0,

∴x₁y₂-y₁x₂=0.

故命題得證.