Question

Answer 1

\(\dfrac{acy-bcx}{c^2}=\dfrac{bcx-abz}{b^2}=\dfrac{abz-acy}{a^2}=\dfrac{acy-bcx+bcx-abz+abz-acy}{a^2+b^2+c^2}=0\)

\(\Rightarrow\left\{{}\begin{matrix}acy-bcx=0\\bcx-abz=0\\abz-acy=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}bx=ay\\cx=az\\cy=bz\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{a}=\dfrac{y}{b}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{y}{b}=\dfrac{z}{c}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)