The many ways in which to express a plane There are many ways to express a plane of $R^3$. I am focusing on two of them.
The first is the cartesian equation $Ax + By + Cz + D = 0$.
The second is to give two direction vectors $u$ and $v$ and a point $P$ of the plane.
My question is: how can I obtain two ortogonal direction vectors $u$ and $v$ and a point $P$ from the cartesian equation $Ax + By + Cz + D = 0$? How can I obtain the cartesian equation from the direction vectors and a point of the plane?
 A: $$(a,b,c)=\vec n$$ is a vector normal to the plane.
From the knowledge of the Cartesian equation, choose the vector among $(0,c,-b), (-c,0,a)$ and $(b,-a,0)$ with the largest norm (do this to avoid degeneracies). This gives you a first vector perpendicular to $\vec n$, let $\vec u$. Then set $\vec v=\vec n\times\vec u$, and you have your second vector.
For the point, you can project the origin orthogonally onto the plane, i.e. find $\lambda$ such that $\lambda(a,b,c)$ fulfills the plane equation. This yields
$$\lambda(a^2+b^2+c^2)+d=0.$$

The converse is easier.
Compute 
$$(a,b,c)=\vec u\times\vec v$$ and expand
$$a(x-x_P)+b(y-y_P)+c(z-z_P)=0.$$
A: From $$Ax+By+Cz+D=0$$
you get first the normal vector to the plane $n=(A,B,C)$.
then you can take
$$u=(0,C,-B)$$
and
$v$ as the vectorial product of $n$ by $u$.
To get the cartesian equation from two vectors $u,v$ and a point $P$, 
$$det(PM,u,v)=0$$
with $M=(x,y,z)$.
A: As other answers point out, you know the normal vector $n=[A,B,C]$. Suppose you have one vector $u\perp n$, $|u|\neq 0$, then clearly you can find $v\perp u$, $v\perp n$ using $v=u\times n$.
So the problem reduces to finding a single nonzero vector perpendicular to $n$.
In 3D, there is no single (non-branching) formula that will do this: see this question.
A: We can write $Ax+By+Cz+D=0$ as:
$$(A,B,C)\cdot (x,y,z) = -D$$
That is, all points $(x,y,z)$ that have the same dot product with $(A,B,C)$, which is $-D$. The shortest vector for which this is the case, is the one that is a parallel to $(A,B,C)$. It means that this vector must be perpendicular to the plane to achieve that. Therefore the vector $\mathbf n$:
$$\mathbf n = \frac{(A,B,C)}{\sqrt{A^2+B^2+C^2}}$$
is the normal vector of the plane with length $1$.
Consequently, we can write the equation as:
$$\mathbf n \cdot (x,y,z) = -\frac{D}{\sqrt{A^2+B^2+C^2}} = d$$
To find a vector $\mathbf P$ in the plane, any vector with $\mathbf n \cdot \mathbf P = d$ will do. That's because the plane consists of all points with the same dot product with $\mathbf n$, which is $d$. We can pick the one that represents the shortest distance:
$$\mathbf P = d\mathbf n = -\frac{D}{A^2+B^2+C^2}(A,B,C)$$
And since $\mathbf n$ is a unit vector, the distance of the origin to the plane must be $|d|=\frac{|D|}{\sqrt{A^2+B^2+C^2}}$. 
To find 2 vectors in the plane, we need 2 independent vectors that are perpendicular to the normal vector $(A,B,C)$. It suffices if the dot product is $0$. Without loss of generality, let's assume that $A\ne 0$. Then we can pick:
$$\mathbf u =(B,-A,0)\quad\text{and}\quad \mathbf v = (C,0,-A)$$
A: Going one direction:
Given $Ax + By + Cz + D = 0$
$(B,-A, 0)$ is a vector in the plane
$(A,B,\frac {A^2 + B^2}{C})$ is a vector in the plane orthogonal to the first.
$(0,0,-\frac {D}{C})$ is a point in the plane
Of course, this is just one way to find orthogonal vectors and a point in the plane.
Going the other direction....
given, vectors in the plane $u,v$ and point in the plane $P$
$u\times v = N$ is normal vector in the plane.
Suppose the components are  $N = (N_x,N_y,N_z)$
then $N_x x + N_y y+ N_z z - N\cdot P = 0$ will be an equation for the plane.
