Question

已知abc为非零数,（a^2+b^2+c^2）(x^2+y^2+z^2)=(ax+by+cz),求证x/a=y/b=z/

Answer 1

因为：(x^2+y^2+z^2)(a^2+b^2+c^2) =(ax)^2+(bx)^2+(cx)^2+(ay)^2+(by)^2+(cy)^2+(az)^2+(bz)^2+(cz)^2

而：(ax+by+cz)^2=(ax)^2+(by)^2+(cz)^2+2abxy+2acxz+2bcyz

则有：(bx)^2+(cx)^2+(ay)^2+(cy)^2+(az)^2+(bz)^2=2abxy+2acxz+2bcyz (ay-bx)^2+(az-cx)^2+(bz-cy)^2=0

固：ay=bx,az=cx,bz=cy

所以：x/a=y/b=z/c