Question

已知a,b不共线,证明c=x₁a+y₁b与d=x₂a+y₂b为共线向量的充要条件为x₁y₂-x₂y₁=0.

Answer 1

证明:设a=(a₁,b₁),b=(a₂,b₂),

∵a,b不共线,∴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.

故命题得证.