在△ABC中,已知向量AB=（x,y）,向量AC=（u,v）求证：△ABC面积S=1/2|xv-yu|

Question

在△ABC中,已知向量AB=（x,y）,向量AC=（u,v）求证：△ABC面积S=1/2|xv-yu|

Answer 1

AB=(x,y),故：|AB|=sqrt(x^2+y^2)
AC=(u,v),故：|AC|=sqrt(u^2+v^2)
AB·AC=(x,y)·(u,v)=xu+yv=|AB|*|AC|*cosA
故：cosA=(xu+yv)/(sqrt(x^2+y^2)sqrt(u^2+v^2))
故：sinA=sqrt(1-(xu+yv)^2/(x^2+y^2)(u^2+v^2))
=sqrt((xv-yu)^2/(x^2+y^2)(u^2+v^2))
=|xv-yu|/(sqrt(x^2+y^2)sqrt(u^2+v^2))
=|xv-yu|/(|AB|*|AC|)
故△ABC的面积：S=(1/2)|AB|*|AC|*sinA
=|xv-yu|/2