How to I go about proving the hyperplane reflection formula? $\displaystyle \tau_v(x)=x-2<x,v>v$
Last edited by dwsmith; Dec 3rd 2011 at 03:21 PM.
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Originally Posted by dwsmith How to I go about proving the hyperplane reflection formula? $\displaystyle \tau_v(x)=-x+2<x,v>v$ Try going to class Dustin! Also he said the error is: \tau_v(x)=x-2<x,v>v
Originally Posted by jonasABRAHA Try going to class Dustin! Also he said the error is: \tau_v(x)=x-2<x,v>v Even with changing that part, do you know how to do it?
tau_v(x) = -(x'-p)=x-p, where x-p = (<v,x>/<v,v>)(v)
Originally Posted by jonasABRAHA tau_v(x) = -(x'-p)=x-p, where x-p = (<v,x>/<v,v>)(v) What is x'? I don't get why you are setting x-p as a projection.
p is a vector in the plan and x its projection (x-p) is its difference, tau_v(x) is its hyperplane reflection which we call x' the different between p and x' -is x'-p