\(\Rightarrow\frac{a\left(bz-cy\right)}{a.a}\)\(=\frac{b\left(cx-az\right)}{b.b}\)\(=\frac{c.\left(ay-bx\right)}{c.c}\)
\(\Rightarrow\frac{abz-acy}{a^2}\)\(=\frac{bcx-baz}{b^2}\)\(=\frac{cay-cbx}{c^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}\)\(=\frac{cay-cbx}{c^2}=\)\(\frac{abz-acy+bcx-abz+cay-cbx}{a^2+b^2+c^2}\)
\(\frac{\left(abz-abz\right)+\left(bcx-cbx\right)+\left(acy-cay\right)}{a^2+b^2+c^2}\)\(=\frac{0+0+0}{a^2+b^2+c^2}=0\)
\(\Rightarrow bz-cy=0;cx-az=0\)
\(\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\frac{y}{b}=\frac{z}{c}\)
\(\Rightarrow cx-az=0\Rightarrow cx=az\Rightarrow\frac{z}{c}=\frac{x}{a}\)
Vậy \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
