Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Oct 25, 2013 #1

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    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. jcsd
  3. Oct 25, 2013 #2
    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

    [tex](a,b,c)\cdot (x,y,z) = d[/tex]

    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

    [tex](a,b,c)\cdot (x,y,z) = 0[/tex]

    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.
     
  4. Nov 3, 2013 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Different Versions of a Plane in R^3 (The Plane!, The Plane!)
Loading...