设F为抛物线y²=4X的焦点,A,B,C为抛物线上三点,若向量FA+向量FB+向量FC=0向量,则|FA|+|FB|+|FC|=?(向量的模)设A,B,C,的横坐标分别为X1,X2,X3则|FA|+|FB|+|FC|=(X1+1)+(X2+1)+(X3+1)=X1+X2+X3+3有因为:
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![设F为抛物线y²=4X的焦点,A,B,C为抛物线上三点,若向量FA+向量FB+向量FC=0向量,则|FA|+|FB|+|FC|=?(向量的模)设A,B,C,的横坐标分别为X1,X2,X3则|FA|+|FB|+|FC|=(X1+1)+(X2+1)+(X3+1)=X1+X2+X3+3有因为:](/uploads/image/z/10914008-32-8.jpg?t=%E8%AE%BEF%E4%B8%BA%E6%8A%9B%E7%89%A9%E7%BA%BFy%26sup2%3B%3D4X%E7%9A%84%E7%84%A6%E7%82%B9%2CA%2CB%2CC%E4%B8%BA%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%B8%8A%E4%B8%89%E7%82%B9%2C%E8%8B%A5%E5%90%91%E9%87%8FFA%2B%E5%90%91%E9%87%8FFB%2B%E5%90%91%E9%87%8FFC%3D0%E5%90%91%E9%87%8F%2C%E5%88%99%7CFA%7C%2B%7CFB%7C%2B%7CFC%7C%3D%3F%EF%BC%88%E5%90%91%E9%87%8F%E7%9A%84%E6%A8%A1%EF%BC%89%E8%AE%BEA%2CB%2CC%2C%E7%9A%84%E6%A8%AA%E5%9D%90%E6%A0%87%E5%88%86%E5%88%AB%E4%B8%BAX1%2CX2%2CX3%E5%88%99%7CFA%7C%2B%7CFB%7C%2B%7CFC%7C%3D%EF%BC%88X1%2B1%EF%BC%89%2B%EF%BC%88X2%2B1%EF%BC%89%2B%EF%BC%88X3%2B1%EF%BC%89%3DX1%2BX2%2BX3%2B3%E6%9C%89%E5%9B%A0%E4%B8%BA%EF%BC%9A)
设F为抛物线y²=4X的焦点,A,B,C为抛物线上三点,若向量FA+向量FB+向量FC=0向量,则|FA|+|FB|+|FC|=?(向量的模)设A,B,C,的横坐标分别为X1,X2,X3则|FA|+|FB|+|FC|=(X1+1)+(X2+1)+(X3+1)=X1+X2+X3+3有因为:
设F为抛物线y²=4X的焦点,A,B,C为抛物线上三点,
若向量FA+向量FB+向量FC=0向量,则|FA|+|FB|+|FC|=?(向量的模)
设A,B,C,的横坐标分别为X1,X2,X3
则|FA|+|FB|+|FC|=(X1+1)+(X2+1)+(X3+1)=X1+X2+X3+3
有因为:向量FA+向量FB+向量FC=0向量,则F(1,0)是△ABC的重心
就是这里不明白,为什么焦点就是重心了.前面都看懂.
设F为抛物线y²=4X的焦点,A,B,C为抛物线上三点,若向量FA+向量FB+向量FC=0向量,则|FA|+|FB|+|FC|=?(向量的模)设A,B,C,的横坐标分别为X1,X2,X3则|FA|+|FB|+|FC|=(X1+1)+(X2+1)+(X3+1)=X1+X2+X3+3有因为:
任意多边形重心到顶点的向量和为0.
设顶点A=(x1,y1,z1)A2=(x2,y2,z2).An=(xn,yn,zn)
重心 M=(x,y,z)
向量MA1+向量MA2+.+向量MAn=(x1-x+x2-x+...+xn-x,y1-y+y2-y+...+yn-y,z1-z+z2-z+...+zn-z)=(x1+...+xn-nx,y1+...+yn-ny,z1+...+zn-nz)
又因为多边形重心坐标=(所有顶点的x的和/n,所有顶点的y的和/n,所有顶点的z的和/n).
所以向量MA1+向量MA2+.+向量MAn=0.
这个是性质..
反过来就是判定,任意多边形到顶点的向量和为0的点是多边形的重心.
运用在本题,是平面上的三边形.
所以若向量FA+向量FB+向量FC=0向量,则F是△ABC的重心.