Finding a linear spanning equation from basis vectors I understand how to find basis vectors from a linear equation, for example, finding the vectors: $\pmatrix{1\\0\\-\frac{3}{2}}, \pmatrix{0\\1\\\frac{1}{2}}$ from the equation $2z-y+3x=0$ but I am struggling to find an intuitive approach to go the other way round, i.e., finding a linear equation from basis vectors. If somebody could explain how this is done with examples I would really appreciate it, thank you.
 A: Two linearly independent vectors in $\mathbb{R^3}$ span a two-dimensional subspace, which is a plane through the origin. So you're looking for the standard  equation in cartesian form:
$$\color{blue}{a}x+\color{blue}{b}y+\color{blue}{c}z=0 \tag{$\star$}$$
where $\color{blue}{(a,b,c)}$ is a normal vector of the plane.
The plane spanned by the two vectors can easily be written in parametric form:
$$\pmatrix{x\\y\\z} = \lambda \pmatrix{1\\0\\-\frac{3}{2}}  + \mu \pmatrix{0\\1\\\frac{1}{2}} \quad\quad \left(\lambda,\mu\in\mathbb{R}\right) \tag{$*$}$$
Eliminating the parameters $\lambda$ and $\mu$ will result in an equation of the form $(\star)$.
If you think about this geometrically, the two spanning vectors are direction vectors of the plane. Their cross product is therefore perpendicular to the plane and can serve as a normal vector.
This leads to a quick method (as already suggested by J. W. Tanner in a comment): take the cross product of the spanning vectors and you immediately get the normal vector $\color{blue}{(a,b,c)}$ to plug into $(\star)$.

The parametric form $(*)$ leads to the following system of equations:
$$\left\{
\begin{array}{l}
\color{purple}{x=\lambda} \\
\color{red}{y = \mu} \\
z = -\frac{3}{2}\color{purple}{\lambda} + \frac{1}{2}\color{red}{\mu}
\end{array}\right.$$
Substitution of the first ($x=\lambda$) and second ($y=\mu$) equation into the third equation, leads to:
$$z = -\frac{3}{2}\color{purple}{x} + \frac{1}{2}\color{red}{y} \iff 2z=-3x+y \iff 3x-y+2z=0$$
In general, elimination of both parameters will be a bit more difficult as we were lucky with some zero coordinates here.
A: If you take the cross product (vector product) of the two basis vectors,
you'll get a vector normal to the plane, which gives the coefficients of the linear equation. 
Using your example, $(1, 0, -\frac 32)\times(0, 1, \frac12)=(\frac32, -\frac12, 1).$ 
The vector $(\frac32,-\frac12,1)$ is then perpendicular to all the vectors in the plane 
spanned by $(1, 0, -\frac 32)$ and $(0, 1, \frac12),$ 
so an equation for the plane is $(\frac32, -\frac12, 1)\cdot(x,y,z)=(\frac32,-\frac12,1)\cdot(1,0,-\frac32)=0$.
A: Take a linear equation $ax+by+cz=0$ and replace both of your vectors $(x,y,z)=(1,0,-\frac{3}{2})$ and $(x,y,z)=(0,1,+\frac{1}{2})$. This will give you two linear equations:
$$ a-\frac{3}{2}c=0$$
$$ b+\frac{1}{2}c=0 $$
The solutions for $a,b,c$ will give you the condition you need. This is equivalent to Tanner approach.
A: It’s exactly the same process that you use to find a basis for the solution set of an equation of the form $ax+by+cz=0$, only the roles of the variables are reversed. In the reverse direction, instead of being given values for $a$, $b$ and $c$, you’re given values for $x$, $y$ and $z$ and want to find the unknowns $a$, $b$, $c$. For your example, this generates the system $$a-\frac32c=0 \\ b+\frac12c=0.$$ The solution is, of course, not unique: the solution space in this case is one-dimensional.  
There’s a simple way to view this in terms of coefficient matrices if you recall that the null space of a matrix is the orthogonal complement of its row space and that the orthogonal complement relation is symmetric. In the forward direction, you would write the coefficients of each equation in the system as the row of a matrix and then compute a basis for the null space of that matrix. Going in the opposite direction, then, you have a basis for the null space of some matrix and want to find a basis for its row space, so write your null space vectors as rows of a matrix and compute its null space normally.  
The above works for any number of equations in any (finite) number of dimensions. If you’re working in $\mathbb R^3$, you can take advantage of cross products to get a solution directly: the two given vectors span a plane through the origin, and its orthogonal complement is spanned by any nonzero vector normal to the plane, which you can find by computing the cross product of the two vectors.
