TITLE: A basis for a subspace of $\mathbb R^3$
QUESTION [5 upvotes]: I have the following question:

Find the basis of the following subspace in $\mathbb R^3$:
  $$2x+4y-3z=0$$

This is what I was given. So what I have tried is to place it in to a matrix $[2,4,-3,0]$ but this was more confusing after getting the matrix $[1,2,-3/2,0]$. This was done to get a leading 1. Now I solved for $x,y,z$.
$$1x + 2y - \frac3 2z = 0$$ from the matrix. Then
$x = \frac3 2z - 2y$
so
$$\left(\frac3 2z-2y, 2y, \frac3 2z\right)$$
Now I have $$(x,y,z) = z(3/2, 0, 3/2) + y(-2, 2, 0) \implies
{(3/2,0,3/2),(-2,2,0)}$$
which has two dimensions.
But this solution that I am giving is wrong. The actual answer according to my text book is ${(3,0,2),(-2,1,0)}$. Can any one please tell me where I messed-up? Thanks.

REPLY [6 votes]: I think there are a little confusions, I'll show it step by step (even the most basic ones), but of course this is the kind of thought that after dealing with math for some time becomes "automatic", so is just a way to understand what is going on. Consider the set:
$$S=\{(x,y,z)\in \Bbb R^3 : 2x+4y-3z=0\}$$
So what this means is that if $(x,y,z) \in S$, then it must be true that $2x+4y-3z =0$, so in truth this is saying that "being an element of $S$ implies satisfying this equation" (of course the converse is also true, but we all know that).
Now, satisfying $2x+4y-3z=0$ implies satisfying:
$$z=\frac{2x+4y}{3}$$
So, since being a point of $S$ implies satyisfying that equation and since satisfying that equation implies that the last coordinate is related to the first two in this way, this chain of implications leads us to the conclusion that being a point  of $S$ implies that it has the following form:
$$p \in S\Longleftrightarrow \ \exists \ x,y\in \Bbb R \ : \ p=\left(x,y,\frac{2x+4y}{3}\right)$$
And because of the operations defined in $\Bbb R^3$ we have that $p$ must be of the form:
$$p = x\left(1,0,\frac{2}{3}\right)+y\left(0,1,\frac{4}{3}\right)$$
This means that being an element of $S$ is the same as being linear combination of those two vectors, so $S$ is the span of those vectors, and since they are linearly independent the definition of basis implies that a basis for $S$ is just the set of those two vectors.