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I assume a 2D coordinate system (system 1), where a point 'A' lies at its origin (0,0), another point 'B' is at 2.4189 distance from point A, at coordinates (1.088,-2.1603) in the 4th Quadrant.

Now if another axis is taken where point B is the origin, now point B will have coordinates (0,0) in this system (system 2), and the axis is rotated in such a way that point A lies straight on the positive y-axis, making its coordinates (0,2.4189). Now in this system, I have a third point C, which has coordinates (4.6332,0.5823). with the available data how can I find the coordinates of point C in system 1.

AN SN1 SN2
System 1 0,0 1.088,-2.1603 x,y
System 2 0,2.4189 0,0 4.6332,0.5823
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What you've described is called a change of basis. Basically, a linear space like $\mathbb{R}^2$ admits an infinite number of basis, or as you called them "axis". Changing basis modifies the components of each vector, or point, but doesn't alter the structure of the space itself.

Basis are changed using matrices, which you can think of as linear functions mapping the space into itself. Let's give it all some formality (Be aware that the following holds for any finite dimension, but for semplicity I will only be using two, as in your example):

$\mathcal{B}$ is the first basis, on which you have two vectors $A=(a_1,a_2)$ and $B = (b_1,b_2)$. On the second basis $\mathcal{B}'$ you know vector $A'=(a_1',a_2')$ and vector $B' = (b_1',b_2')$. To find the change of basis matrix you need to solve the following system for $c_{i,j}$

$$ \left\{ \begin{array}{c} a_1' = c_{1,1}\cdot a_1 + c_{1,2}\cdot a_2 \\ b_1' = c_{1,1}\cdot b_1 + c_{1,2} \cdot b_2\\ a_2' = c_{2,1}\cdot a_1 + c_{2,2}\cdot a_2 \\ b_2' = c_{2,1}\cdot b_1 + c_{2,2} \cdot b_2 \end{array} \right. $$

Now you can actually build the change of basis matrix $C = \{c_{i,j}\}$. To find where any other vector $u\in\mathbb{R}^2$ is being mapped to, you just need to multiply it by the matrix $C$ and you have your answer.

If you want to go a little further, note that the matrix $C$ is square and with non-zero determinant, and therefore invertible. If you multiply the vectors in the second basis $A',B'$ by the inverse $C^{-1}$ you will get back the vectors in the first basis $A,B$.

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  • $\begingroup$ Thanks a lot for your explanation. It was really helpful. $\endgroup$ Apr 8, 2021 at 16:17
  • $\begingroup$ In change of basis I need to keep the origin unchanged? In this example of mine I have changed the origin as well. The new axis not only is different from the angle point of view but also from the origin point of view. $\endgroup$ Apr 9, 2021 at 7:09
  • $\begingroup$ Here things get a little tricky. As a general rule, linear maps, and therefore matrices, keep the origin unchanged, or more formally they ALWAYS send the zero vector into the zero vector. What you have described is a TRANSLATION, meaning every vector $v\in\mathbb{R}^2$ is mapped into $v' \rightarrow v+\mu$, where $\mu$ is the translation vector. Such a map is NOT linear, as the translation of the zero vector is not a the zero vector. Thus, translations can not be represented by matrices $\endgroup$ Apr 9, 2021 at 17:39
  • $\begingroup$ However, this is not a problem. Since we are dealing with a ROTATION (matrices on $\mathbb{R}^2$ make the vector rotate around the origin) and a TRANSLATION, we just have to apply one after the other. Given a vector $v\in\mathbb{R^2}$, to calculate where it's map we just do $Cv+\mu$. $\endgroup$ Apr 9, 2021 at 17:46
  • $\begingroup$ There are in fact some ways to deal with translation from a matricial perspective, but that would require to move upwwards to the $\mathbb{R}^3$ space. If you interested in the most general treatment about spatial transformations, the combination of rotations and translations and actually something more, google Galileian group. $\endgroup$ Apr 9, 2021 at 17:52

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