Finding Isometry between Two Parametric Lines in R^3

lollikx89
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Homework Statement


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.


Homework Equations


What i have to find is a 3x3 matrix which 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.


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.
 
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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
 
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.
 
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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