Ta có : \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)= \(\frac{bza-cya}{a^2}=\frac{cxb-âzb}{b^2}=\frac{ayc-bxc}{c^2}\)
= \(\frac{bza-cya+cxb-azb+ayc-bxc}{a^2+b^2+c^2}\)\(=\frac{0}{a^2+b^2+c^2}=0\)
Suy ra : bz - cy = 0 \(\Rightarrow\) bz= cy \(\Rightarrow\) \(\frac{z}{c}=\frac{y}{b}\) (1)
cx - az = 0 \(\Rightarrow\) cx = az \(\Rightarrow\) \(\frac{x}{a}=\frac{z}{c}\) (2)
ay - bx = 0 \(\Rightarrow\) ay = bx \(\Rightarrow\)\(\frac{y}{b}=\frac{x}{a}\) (3)
Từ (1) , (2) và ( 3) \(\Rightarrow\)\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) (điều phải chứng minh )
