Write the plane in vector equation form

Question

$2x+4y-4z=4$ is the point normal equation form for the plane.

How do we write the plane in vector equation form, as $$(x,y,z)=(*,*,*)+t_1(*,*,*)+t_2(*,*,*)$$

Answer 1

The plane equation can be rewritten as: $$x+2y-2z=2.$$ We can set $y=s,z=t$, so $x=2-2s+2t$. So, we have the parametric system: $$\left\{\begin{array}{l} x=2-2s+2t\\ y=s\\ z=t \end{array} \right.\iff \left\{\begin{array}{l} x=2-2s+2t\\ y=0+1s+0t\\ z=0+0s+1t \end{array} \right. $$

Thus, the system can be written in the vector form: $$(x,y,z) = (2,0,0)+s\cdot (-2,1,0) + t\cdot (2,0,1), \quad s,t \in \mathbb R.$$

Answer 2

The plane equation can be written as a scalar product ${\bf v}{\bf n}^t = k$ where $k$ is a scalar constant, ${\bf v} = (x,y,z)$ and $\bf n$ is a vector which is normal to the plane, in this case $(2,4,-4)$. Any change in $\bf v$ which would cause a change in the scalar product can not be allowed, since then we could not have equality any longer. So if we are at a point on our plane, then we can only move in a direction which gives ${\bf (\Delta v)n}^t = 0$.

So which vectors give $2x+4y-4z = 0$, one such vector is $(0,1,-1)$ and another is $(2,0,1)$. We could just choose those as directions for the parametrization, then what is left is to find one point on the plane which is the term which is first (without any $t_{(.)}$).