Solution space of system of two homogeneous linear equations

Question

I have the following problem: Show that any line through the origin in $\mathbb{R^3}$ is the solution space of a system of two homogeneous linear equations.

I have tried the following: let the line be described in the following way: $$X = \alpha(x_1, x_2, x_3).$$ Then the solution space is given by $(y_1, y_2, y_3) = (\alpha x_1, \alpha x_2, \alpha x_3)$, giving the set of three equations $$y_1 = \alpha x_1 $$ $$y_2 = \alpha x_2 $$ $$y_3 = \alpha x_3.$$

This is a set of three equations, not two, and I don't see what I'm doing wrong. There is something I'm not understanding I think. Any help would be appreciated. Thank you.

Answer 1

No, what it is telling is following:

If $L$ is any line passing through origin, then there exist two homogeneous linear equations say $a_1x+b_1y+c_1z=0$ and $a_2x+b_2+c_2z=0$ such that the solution space of the two linear equations will be the line $L.$

Let $L$ be a line passing through the origin. So $L$ is the span of some nonzero element, say $(x_0,y_0,z_0)\implies L=\{(\alpha x_0,\alpha y_0,\alpha z_0)): \alpha \in \mathbb{R}^3\}$. Now you can find the equation of the plane which contains the line $L$ and passes through the point $(x_1,y_1,z_1) $ which is not on the line (See here). Again you can find one more plane which contains the line $L$ and passes through the point $(x_2,y_2,z_2) $ which is not on the line. Here you can do this because the dimension of $\mathbb{R}^3$ is $3$ so there exists a basis containing three elements. And, hence you got two linear homogeneous equations, plane $1$ and plane $2$ whose solution space is the entire line $L$.