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

1. Oct 25, 2013

### 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)?

2. Oct 25, 2013

### 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.

3. Nov 3, 2013

### 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.

Last edited: Nov 3, 2013