A basis for a subspace of $\mathbb R^3$ 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.
 A: 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.
A: I'll follow your method as closely as I can to make it clearer to you.
We get an equation $$2x+4y-3z=0 \implies x+2y-\boxed{\frac3 2z}=0 \implies x = \frac3 2z-2y$$
so that
$$(x,y,z)=\left(\frac 3 2 z-2y,y,z\right) = z\left(\frac 3 2,0,1\right) + y\left(-2,1,0\right)$$

Note that there's no reason why the basis you get and the basis they get should be the same, because there are infinitely many bases spanning the same space. For example, for any basis $\mathbf{x},\mathbf{y}$ the vectors $\mathbf{x+y},\mathbf{x-y}$ are a basis too.
A: You can get a basis quickly, as follows. You have $$2x+4y-3z=0$$.
This means, $$z=\frac{2x+4y}3$$
So if $\vec x$ solves the above, $$\vec x=\left(x,y,\frac 2 3 x+\frac 4 3 y\right)$$
That is, $$\vec x =x\left(1,0,\frac 23\right)+y\left(0,1,\frac 43\right)$$
So you can choose your basis to be $\{(3,0,2),(0,3,4)\}$ upon scaling.
In general, if you're working on $\Bbb R^3$; you know $ax+by+cz=0$ will be a subspace of dimension two (a plane through the origin), so it suffices to find two linearly independent vectors that satisfy the equation. To that end, make a coordinate vanish, say $x=0$, and find what $y,z$ may be. For example, in you case we get $4y-3z=0$, so we can take $y=3,z=4$, to get the first vector to be $(0,3,4)$. Now make $y=0$. You get $2x-3z=0$, so this suggests taking $x=3,z=2$. This gives the basis $$\{(0,3,4),(3,0,2)\}$$ as above. Note that if we chose to make say $z=0$; we could put $x=-2,y=1$ to get $$\{(-2,1,0),(3,0,2)\}$$
This works nicely because we have one equation, but wouldn't work so swimmingly if we had something of the form $$\begin{cases} 2x+3y-2z=0\\-x+y+3z=0\end{cases}$$
In fact, setting any coordinate to $0$ in the above gives nothing. However, setting say $z=1$ gives $x=11/5$ and $y=-4/5$, which correctly gives the solution set is $$\langle (11,-4,5)\rangle$$
A: Note that you should have divided $-3z$ by $2$ to obtain: $x=-2y+\dfrac{3}{2}z$, which yields:
$$\left[ \begin{array}{c}
x \\
y \\
z
\end{array} \right]
=
\left[ \begin{array}{c}
-2y+\dfrac{3}{2}z \\
y \\
z
\end{array} \right]
=
\left[ \begin{array}{c}
-2y \\
y \\
0
\end{array} \right]
+
\left[ \begin{array}{c}
\dfrac{3}{2}z \\
0 \\
z
\end{array} \right]
=
y\left[ \begin{array}{c}
-2 \\
1 \\
0
\end{array} \right]
+
\dfrac{1}{2}z\left[ \begin{array}{c}
3 \\
0 \\
2
\end{array} \right]
$$
where I factored out the $1/2$ to avoid fractions (this step wasn't necessary, but makes the answer prettier). Thus, we obtain the desired basis.
A: First you dont need to put it in a matrix, as it is only one equation, you can solve right away.  For gettin the generators of that subspace all you need to do is put a variable in function of the other ones, be $3z-4y = 2x$ for example, then you have that $x = \frac{3}{2} z-2y$ replacing in a vector you get $(\frac{3}{2} z-2y,y,z)$ separating the sum as x and y are not in function of each other $z(\frac{3}{2} , 0, 1) + y(-2,1,0)$ and multiplying the firt by 2 as as it doesnt change the solution, they're linearly independant you get a base {${(3,0,2)(-2,1,0)}$}, and the dimension of this is 2.
Your mistake was to put 2y and  1/3 z in the second and third coordinates, as this only affect X (the first coordinate), there it should be y and z
A: Its not 1x+2y-1/3z=0, but 1x+2y-1/2z=0. That is your main mistake. You can also avoid fractures by looking at vector (2x,2y,2z)=(3z-4y,2y,2z)=z.(3,0,2)+y.(-4,2,0)
