向量和三角形的问题
△ABC的垂心为H,
向量HA*HB=HB*HC=HC*HA。①
设a=OA=(x1,y1),b=OB=(x2,y2),c=OC=(x3,y3),OH=(x,y)。
则向量HA=(x1-x,y1-y),HB=(x2-x,y2-y),HC=(x3-x,y3-y),
①化为(x1-x)(x2-x)+(y1-y)(y2-y)=(x2-x)(x3-x)+(y2-y)(y3-y)
=(x3-x)(x1-x)+(y3-y)(y1-y),
∴x1x2-x(x1+x2)+y1y2-y(y1+y2)=x2x3-x(x2+x3)+y2y3-y(y2+y3)
=x1x3-x(x1+x3)+y1y3-y(y1+y...全部
△ABC的垂心为H,
向量HA*HB=HB*HC=HC*HA。①
设a=OA=(x1,y1),b=OB=(x2,y2),c=OC=(x3,y3),OH=(x,y)。
则向量HA=(x1-x,y1-y),HB=(x2-x,y2-y),HC=(x3-x,y3-y),
①化为(x1-x)(x2-x)+(y1-y)(y2-y)=(x2-x)(x3-x)+(y2-y)(y3-y)
=(x3-x)(x1-x)+(y3-y)(y1-y),
∴x1x2-x(x1+x2)+y1y2-y(y1+y2)=x2x3-x(x2+x3)+y2y3-y(y2+y3)
=x1x3-x(x1+x3)+y1y3-y(y1+y3),
∴{x(x3-x1)+y(y3-y1)=x2(x3-x1)+y2(y3-y1),
{x(x1-x2)+y(y1-y2)=x3(x1-x2)+y3(y1-y2)。
解上述关于x,y的方程组就得到点H的坐标。
我无法用a,b,c表示OH。
。收起