求解一道高中数学题

显示全部楼层 · 2013-1-15 15:25:10

设AM延长线与BC交于D点。即有D是BC的中点。那么有向量PD=1/2PB+1/2PC又M是重心，则有AM=2MD故有PM=1/3PA+2/3PD即有PM=1/3PA+2/3（1/2PB+1/2PC）=1/3PA+1/3PB+1/3PC即有X=1/3，Y=1/3，Z=1/3X+Y-Z=1/3选择A...

千问 · 2013-1-15 15:25:10

PM=PA+AM
=PA+2/3(AB+1/2BC)
=PA+2/3(AP+PB+1/2(BP+PC))
=1/3(PA+PB+PD)所以X=Y=Z=1/3X+Y-Z=1/3所以选A...

千问 · 2013-1-15 15:25:10

PM=1/3PA+1/3PB+1/3PC答案：A...

千问 · 2013-1-15 15:25:10

应该是 B...........