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Reflection across a Plane

  1. Aug 10, 2010 #1
    Suppose you have the plane given by

    [tex]\bold n \cdot (\bold r-\bold r_0)=0[/tex]

    where

    [tex]\bold n[/tex] is the normal vector to the plane which passes through the point [tex] \bold r_0[/tex].

    What is the reflection [tex]x'[/tex] of a point [tex]x[/tex] across this plane?
     
  2. jcsd
  3. Aug 11, 2010 #2

    quasar987

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    Consider [itex]\mathbf{r}'_1, \mathbf{r}'_2[/itex] two points in the plane such that the vectors [itex]\mathbf{r}_1:=\mathbf{r}_1-\mathbf{r}_0, \mathbf{r}_2:=\mathbf{r}_2-\mathbf{r}_0[/itex] are linearly independant. (Think of it this way: the vectors r_1,r_2 based at r_0 form a basis for the plane.) Then any point [itex]\mathbf{x}[/itex] of R^3 corresponds to a triple (a,b,c), where

    [tex]\mathbf{x}=\mathbf{r}_0+a\mathbf{r}_1+b\mathbf{r}_2+c\mathbf{n}[/tex]

    The reflection of x with respect to the plane is the point x' corresponding to the triple (a,b,-c). I.e.

    [tex]\mathbf{x}'=\mathbf{r}_0+a\mathbf{r}_1+b\mathbf{r}_2-c\mathbf{n}[/tex]
     
  4. Aug 11, 2010 #3
    Thanks quasar987!
     
  5. Aug 11, 2010 #4
    you can take this a step further, and get rid of the coordinates.

    (x-r0) . n = a (r'1 - r0) . n + b (r'2 - r0) . n + c n . n = c

    and
    x' = r0 + a r1 + b r2 - c n
    =r0 + a r1 + b r2 + c n - 2 c n
    = x - 2 c n
    = x - 2 ((x-r0) . n) n

    /assuming we normalized n . n = 1
     
  6. Aug 15, 2010 #5
    Thanks qbert!
     
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