Question

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

Answer 1

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.(t∈R),∴|c|²=x²+y²=(1/2)sint+cost+(3/2)=(√5/2)sin(t+w)+(3/2).(tanw=2).即|c|²=(√5/2)sin(t+w)+(3/2).∴|c|²max=(3+√5)/2=(6+2√5)/4=(1+√5)²/4.∴|c|max=(1+√5)/2.