Question

[Nguyễn Công Anh]

Nếu \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)thì \(\left(x^2+y^2+x^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

Answer 1

[Mr Lazy]

\(\Leftrightarrow\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2=0\)