I need to confirm if the Euclidean distance between two points still remains the same if the axis is rotated at a certain angle (two dimensional)? for example, point A and point B have a distance of 3 units in coordinate system 1. If the coordinate system is rotated at an angle theta, in this new system (system 2) will have the same Euclidean distance of 3 units or not?
I am trying to solve a problem that has both translation and rotation of the axis. So I can say I will have three-axis in this process, first is the original, second after translation, and third after the rotation of this second axis.
Yes, if you can perform such distribution in spherical form. It will be called antipodal symmetric distribution. Also by parallelogram law your arguments are correct, with no worries. See details here : https://au.mathworks.com/help/symbolic/rotation-matrix-and-transformation-matrix.html
Also http://motion.cs.illinois.edu/RoboticSystems/3DRotations.html
Muhammad Ali Thanks for your comment. The one I am working with is two-dimensional.
Dear Sohaib Bin Altaf Khattak
The answer is yes.
The distance between the points preserved after axis rotation by angle
Proof: Assume that A(a,b) and B(c,d) are given by
d²(A,B)=(c-a)²+(d-b)²
After rotation by angle θ,
the new coordinates of A and B are
A₁( acos(θ)+bsin(θ),-asin(θ)+bcos(θ))
B₁( ccos(θ)+dsin(θ),-csin(θ)+dcos(θ))
d²(A₁,B₁)=((ccos(θ)+dsin(θ))-(acos(θ)+bsin(θ)))²
+((-csin(θ)+dcos(θ))-(-asin(θ)+bcos(θ)))²
= (-2ac-2bd+a²+b²+c²+d²)(cos²θ+sin²θ)
=(c-a)²+(d-b)²=d²(A,B).
Best regards
If you use polar coordinates to identify points, then a rotation of the coordinate system is easily accomplished by subtracting the rotation angle from each point. You can think of the distance between A and B as a side of a triangle with the third vertex at the origin.
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