Different Versions of a Plane in R^3 (The Plane, The Plane)

[WWGD]

Different "Versions" of a Plane in R^3 (The Plane!, The Plane!)

Hi, All:

The two most common expressions , AFAIK, for describing a plane in $R^3$ are:

i) Writing the plane in the form : ax+by+cz=d

ii) Giving a basis {v₁, v₂} for the plane.

I know how to produce a basis to go from version i) to version ii) -- basically by using techniques of Gaussian elimination . I know of a method to produce a plane from a pair of basis elements, but I can't see how it works ; we do a cross-product of the basis vectors. I thought this cross-product would just produce a vector normal to the given {v₁ , v₂} , but somehow it generates an equation of the type in i). Does anyone understand why/how this formula works, or know some other nice way to go from version ii), i.e., from a basis, to the form in i)?

 

[R136a1]

First of all, what you did is not completely accurate. In (ii), you should have a basis and a point on the plane.

Anyway, for your question. You should know that the first equation can be written as

(a,b,c)\cdot (x,y,z) = d

Now, to find a basis for the plane, you need to find two linearly independent points on the plane going through the origin and parallel to the original plane. This yields the equation

(a,b,c)\cdot (x,y,z) = 0

So, if you are given a basis, then you are given two points whose dot product with $(a,b,c)$ is $0$. So it is perpendicular to $(a,b,c)$. To find a vector orthogonal to the basis vectors is usually done with the cross product. So the cross product gives one possibility of $(a,b,c)$. To determine $d$, you need to use the point on the plane.

 

[WWGD]

Ah, yeah, sorry I forgot to include the most important part; clearly, like you said, the method for obtaining a basis from the equation of the plane can be reversed to get the equation of the plane from a pair of basis vectors; that I get. It comes down to going from affine space to vector space (going thru the origin or not) and back.

BUT**I forgot to say that I'm dealing here with a rotating system of planes (a.k.a, moving frame), i.e., a "plane distribution" , i.e., an assignment of planes at each point --at each tangent space, actually--in coordinates $$ (r, \theta, z)$$ and this coordinate system rotates about the XY-plane, so that, at a fixed point $$(x_0,y_0,z_0)$$:

1)The vector $$\partial /\partial z $$ : has a fixed direction pointing upwards.

2)The vector $$\partial /\partial r $$ : is in the direction of the line from $$(0,0,z_0) $$ to $$(x_0,y_0,z_0)$$, i.e., $$ \partial /\partial r $$ is in the plane $$z=z_0$$.

3)The vector $$ \partial/ \partial \theta $$ is in the $$z=z_0 $$ -plane also, and it is perpendicular to

$$ \partial / \partial r $$ in that plane.

So we have some moving frames here, which I think changes the layout.