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

Unique set of vectors normal to a hyperplane

  1. Oct 14, 2012 #1
    Let's say we have a point-normal representation of a space:
    [tex] n \cdot P_{0}P = 0 [/tex] where n is a vector [itex]<a_{1},a_{2}...a_{n}>[/itex] and [itex]P_{0}[/itex] is a point through which the space passes and P is the set of all points contained in the space.

    In [itex]ℝ^{2}[/itex], the point-normal representation defines a line.
    In [itex]ℝ^{3}[/itex], the point-normal representation defines a plane.
    In [itex]ℝ^{n}[/itex], the point-normal representation defines a hyperplane, or (n-1) dimensional affine space.

    It can be shown that this affine space can be represented in component form as:
    [tex] a_{1}x_{1} + a_{2}x_{2} + ... + a_{n}x_{n} = b [/tex]
    where b is a constant.

    My question is essentially asking about the converse of the representation done above: Can we go from the component-wise representation to the point-normal representation?
    In other words, can we show that the only vectors normal to the space determined by [tex] a_{1}x_{1} + a_{2}x_{2} + ... + a_{n}x_{n} = b [/tex] are vectors of the form
    [itex]t<a_{1},a_{2}...a_{n}>[/itex] where t is a scalar parameter?

    BiP
     
  2. jcsd
  3. Oct 15, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey Bipolarity.

    You could use the identity that <tx,y> = t<x,y> for general inner products and if <x,y> = 0 then <tx,y> = t<x,y> = 0 for all x and y in the inner product space.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook