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Find isometry

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data
    i was given two parametric equations of two lines in R^3 and asks me to find the isometry between one line and the other knowing that point (a,b,c) of first line is mapped in point (a',b',c') of second line.


    2. Relevant equations
    What i have to find is a 3x3 matrix wich maps the point of the first line into the point of the second line maintaining the distance between two points of the first line and their images on the second line.


    3. The attempt at a solution
    Trying to determine such a map caused me some troubles. I thought i should multiply the generic point of the first line with a general 3x3 matrix and impose that the image of this point is at the same distance from (a',b',c') that the first generic point from (a,b,c) but this method obviously does not work because i have a system with 9 elements to find ( the 9 elements of the matrix). For sure i'm doing something wrong but i couldn't find what. Thanks in advance for any help.
     
  2. jcsd
  3. Dec 16, 2011 #2

    Office_Shredder

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    It might help to think a bit more geometrically. You have two tasks: translate one point to another point (hopefully you have that part handled), and make one line point in the same direction as another line. You don't really need to worry about arbitrary points on the line, just what direction your line is pointing in
     
  4. Dec 16, 2011 #3
    Ok, so i consider the two points i was given and the two directions identified from the two lines. What i should do is a translation and a rotation right? So i should need a matrix 4x4 and composing translation with rotation. My problem is to identify that matrix because i need the angle of rotation for each axe to determine it right? I'm a bit confused about founding the matrix.
     
  5. Dec 16, 2011 #4

    chiro

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    If you are doing a transformation from R^3 to R^3 then given that a line has the equation l(t) = at + (1-t)b for points a and b on the line should translate naturally to l'(t) = a't + (1-t)b'.

    Given this information, does this help you?
     
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