已知平面向量a,b,c,若a=(1,0),b=(1,1),且(a-c)(b-c)=0,则c的最大值等于

可设向量c=(x,y).则向量a-c=(1,0)-(x,y)=(1-x,-y),b-c=(1,1)-(x,y)=(1-x,1-y).由题设可得(1-x,-y)·(1-x,1-y)=0.===>(1-x)²-y(1-y)=0.===>(x-1)²+[y-(1/2)]²=1/4.换元,可设x=1+(cost)/2.y=(1+sint)/2...