Basis for the subspace constituted by the intersection of two planes

Show that the intersection of the planes $4x+5y+6z = 0$ and $x+2y+5z = 0$ constitute a subspace of $\mathbb R^3$. Find a basis for this subspace.

I got the product of normal vectors as $\langle 13,-14,3\rangle$.

If I try to substitute $z=0$, I get $x$ and $y$ also as $0$. If I solve the given two equations as homogeneous linear system, I get $x=13z/3$, $y=14z/3$.

How to proceed after this and prove this intersection line constitute a subspace in $\mathbb R^3$ and finding basis for this kind of vectors with dependent variables?

Solution 1:

You have shown that the intersection consists of vectors of the form $(x, y, z)$ such that $x = 13z/3$ and $y = 14z/3$. That is, points of the form $$\left(\frac{13}{3}z, \frac{14}{3}z, z\right) = z\left(\frac{13}{3}, \frac{14}{3}, 1\right)$$ for $z \in \Bbb{R}$. That is, you have shown that the intersection is precisely $$\left\{z\left(\frac{13}{3}, \frac{14}{3}, 1\right) : z \in \Bbb{R}\right\} = \operatorname{span}\left\{\left(\frac{13}{3}, \frac{14}{3}, 1\right)\right\}.$$ This shows two things: the intersection is the span of a subset of $\Bbb{R}^3$, which makes it automatically a subspace, and that this subspace is spanned by a set containing a single non-zero vector (which makes it automatically linearly independent), so we also have a basis.