My question is about converting rectangular equation of a $3D$ plane in $\Bbb{R}^4$ to its parametric form, like that:
Given the plane $M = \{ (x,y,z,w)\in \Bbb{R}^4 \mid 2x-2y-6z=1 \}$, Find its parametric equation.
By subtitutions we get:
$w =r, z = s, y = t, x = \frac{1}{2}+t+3s $
so: $ M=\{ (\frac{1}{2},0,0,0)+t(1,1,0,0)+s(3,0,1,0)+r(0,0,0,1) \} $
My question is, should i throw out the $r$ component that represents the value of $w$ because $M$ is a $3D$ plane? or not - since $w = r$ and for each $r\in R$ the plane remains the same.
In general, could someone tell me which of the following is the correct answer and why?
$ M=\{ (\frac{1}{2},0,0,0)+t(1,1,0,0)+s(3,0,1,0)+r(0,0,0,1) \} $
$ M=\{ (\frac{1}{2},0,0,0)+t(1,1,0,0)+s(3,0,1,0)+r(0,0,0,0) \} $
$ M=\{ (\frac{1}{2},0,0,0)+t(1,1,0,0)+s(3,0,1,0) \} $
All the above planes seem to me the same, because the value of $w$ has no effect on the plane itself $(M = \{ (x,y,z,w)\in \Bbb{R}^4 \mid 2x-2y-6z=1 \})$
By the way, when I converted back the above parametric equations to their rectangular (Cartesian) form, I got same results for all of them.
So, how will the parametric equation looks like?
Thanks for help!!
Since $M$ is three-dimensional, its parameterization must have three free variables. There’s nothing to “throw out.” The $w$-coordinate’s being unconstrained is not at all the same thing as saying that “it has no effect.” You still need a way to generate points with nonzero $w$-coordinates since any value of $w$ will satisfy the defining equation.
As for the other parameterizations, $r(0,0,0,0)$ adds literally nothing to the second one, so it effectively has only two free variables. If you got the same result when converting each of these parameterizations into implicit Cartesian form, then you did something very wrong, since the latter two describe a two-dimensional plane, not a three-dimensional hyperplane.