Converting Plane Equation from Cartesian Form to Parametric Form.

Question

I need to convert a plane's equation from Cartesian form to Parametric form. For example:

                   2x-y+6z=0

to: the vectors

        (a, b, c) + s(e, f, g) + t(h, i, j)

So basically, my question is: how do I find the $a,b,c,s,e,f,g,t,h,i,j$ and what's the logic behind the conversion.

I found a solution from http://www.easy-math.net/transforming-between-plane-forms/ and attached the screen shot of it as an image. I'm wondering if their conversion formula is correct.

Answer 1

Your choices for $a,b,c,d,e,f,g,h,i$ are fairly wide open.

You need to find a point on your plane. Any point will do, there are infintely many of them.

That point will define $(a,b,c)$ Now, you need to find any two independent (non-parallel) vectors that lie in the plane. They might be perpendicular, but they don't have to be.

In the example above. $2x - y + 6z = 0$

$(0,0,0)$ is a point in the plane.

$2\cdot 0 - 1 \cdot 0 + 6\cdot 0 = 0$

$(1,2,0)$ is a vector in the plane.

That is, from any point in the plane, if we increase $x$ by $1$ and $y$ by $2$, we are still in the plane.

$(0,6,1)$ is also a vector in the plane.

And they are independent. There is no way to combine the two vectors such that

$(1,2,0)s + (0,6,1)t = \mathbf 0$ unless $s = t = 0$

$s$ and $t$ are your parameters and they stay $s$ and $t$

$(x,y,z) = (0,0,0) + (1,2,0)s + (0,6,1)t$ would be one representation.

As I said above, this is not unique. we could choose a different point in the plane, and we could choose different vectors.

Answer 2

The short answer is to find three points on the plane, select one as a "source" and have the other two as vectors.

For $2x - y + 6z = 0$, three easy points to select are $(0, 0, 0), (3, -6, 0),$ and $(0, 6, 1)$. Call these, respectively, $P, \ Q,$ and $R$. Form vectors $\vec{PQ}$ and $\vec{PR}$ which are, respectively, $(3, -6, 0)$ and $(0, 6, 1)$. These form the directions for parameters $s$ and $t$. We can thus parameterize this plane as

$$ \mathbf{r}(s, t) = (3, -6, 0)s + (0, 6, 1)t $$

This, of course, is just one example of doing this. Remember that three non-colinear points form a plane.