A solution to the system of equations
$ax + by +cz = 0$
$a^2x+b^2y+c^2z=0$ is
Ans: $x=\frac{k (b-c)}{a}$ $y= \frac{k (c-a)}{b}$ $z=\frac{k (a-b)} {c}$ where $k$ is an arbitrary constant.
A system of two equations in three variables is supposed to given an equation that in one independent variable. How do I reconcile this fact to the given solution and how do I reach this solution ?
On
solve your first equation for $z$ then we get $$z=-\frac{a}{c}x-\frac{b}{c}y$$ plugging this in your second equation we obtain $$a^2x+b^2y+c^2\left(-\frac{a}{c}x-\frac{b}{c}y\right)=0$$ solve this for $x$ and set $$y=k$$ simplifying the equation above we obtain $$x(a^2-ac)=(bc-b^2)y$$ so $$x=\frac{(bc-b^2)y}{a^2-ac}$$ if $$a^2-ac\neq 0$$ with $$y=k$$ we have $$z=-\frac{a}{c}\left(\frac{(bc-b^2)y}{a^2-ac}\right)-\frac{b}{c}y$$ we Can write $$\left\{ x={\frac {cz \left( -c+b \right) }{ \left( a-b \right) a}},y= -{\frac {c \left( a-c \right) z}{b \left( a-b \right) }},z=z \right\} $$
They are equations of two planes and we need to find an equation of the common line of these planes.
Thus, with assumption that they are not parallel planes we need to prove that $$(a,b,c)\perp\left(\frac{b-c}{a},\frac{c-a}{b},\frac{a-b}{c}\right),$$ which is $$a\cdot\frac{b-c}{a}+b\cdot\frac{c-a}{b}+c\cdot\frac{a-b}{c}=0$$ and $$(a^2,b^2,c^2)\perp\left(\frac{b-c}{a},\frac{c-a}{b},\frac{a-b}{c}\right),$$which is $$a^2\cdot\frac{b-c}{a}+b^2\cdot\frac{c-a}{b}+c^2\cdot\frac{a-b}{c}=0,$$ which is obvious.