a·c=b·c,即:(a-b)·c=0
设c=(x,y),a-b=(√3-1,-1-√3)
即:(√3-1)x=(√3+1)y
即:y=(2-√3)x
又:x^2+y^2=2
即:x^2+(7-4√3)x^2=2
即:x^2=1/(4-2√3)=(4+2√3)/4
故:x=(√3+1)/2或-(√3+1)/2
y=(√3-1)/2或(1-√3)/2
即:c=((√3+1)/2,(√3-1)/2)或
c=(-(√3+1)/2,(1-√3)/2)
a·c=b·c,即:(a-b)·c=0
设c=(x,y),a-b=(√3-1,-1-√3)
即:(√3-1)x=(√3+1)y
即:y=(2-√3)x
又:x^2+y^2=2
即:x^2+(7-4√3)x^2=2
即:x^2=1/(4-2√3)=(4+2√3)/4
故:x=(√3+1)/2或-(√3+1)/2
y=(√3-1)/2或(1-√3)/2
即:c=((√3+1)/2,(√3-1)/2)或
c=(-(√3+1)/2,(1-√3)/2)