On the Invariance of the Distances Between Points in a Plane under Rotations of the Coordinate System

 

Hi there. As the title already suggests, this post is going to be a little academic. I found the following question among the challenge questions of the "Vectors" section of an “University Physics” book [1] and wanted to publish its solution, just because I liked it. Actually, high school math should be enough to follow the solution. I will try to simplify it anyways, as much as I can (so that I can also understand it afterwards).

Question: Distances between points in a plane do not change when a coordinate system is rotated. In other words, the magnitude of a vector is invariant under rotations of the coordinate system. Suppose a coordinate system S (red) is rotated about its origin by angle φ to become a new coordinate system S' (blue), as shown in the following figure.

Image Source: University Physics Vol. 1 [1]

A point in a plane has coordinates (x, y) in S and coordinates (x', y') in S'.

(a) Show that, during the transformation of rotation, the coordinates in S' are expressed in terms of the coordinates in S by the following relations:

Remark: The above formula is only correct, if the angle φ is negative (clockwise). In the image, the angle shows a positive direction. I have still tried to stick to the original formula given in the book.

At this point, a good mathematician would express the above equations with a 2D rotation matrix.

This is a much deeper subject, when generalized to 3D. I'm going to publish another post on this in the future.

(b) Show that the distance of point P to the origin is invariant under rotations of the coordinate system. Here, you have to show that: 

(c) Show that the distance between points P and Q is invariant under rotations of the coordinate system. Here, you have to show that:

Preliminary Information

First of all, since there is a rotation in the question, it is possible to simplify the operations by switching to polar coordinate system. The coordinate system S in the image is called Cartesian coordinate system (CCS), where the perpendicular distances of a point in the plane are given in (x, y) pairs. In addition, the axes can be rotated, as shown in the image, and they also don't need to be always perpendicular to each other (also see: Orthogonal Coordinates). Skewed coordinate systems are also possible. The only condition of existence of such coordinate systems here is, that the axes are not overlapping, i.e. the angle between them is not zero. In case the angle is zero, two dimensions collapse to one (N dimensions are likewise reduced to N-1), and we get the number line, which we know from the elementary school: The number line is a one-dimensional coordinate system. The term "Cartesian" is meaningless, as there is orthogonality of a single vector. 

In polar coordinate system (PCS), a point is expressed by the pairs (r, θ), where r is the distance from the origin and θ is the positive (counterclockwise) angle between the line segment and the horizontal axis.

Note: In general, to express any point in an N-dimensional space, N terms are required regardless of the coordinate system chosen. E.g., in 3D, the triples (x, y, z) are used in the cartesian coordinate system, (r, θ, φ) in the spherical coordinate system or (ρ, φ, z) in the cylindrical coordinate system.

Therefore, I can express the problem in PCS as follows.

Representation of the problem in PCS

As it can be seen, the position of the points P or Q has not changed. We just express them differently. As an analogy, the concept of 'book' exists independently of language, but it's obviously being expressed in different words in different languages. If I'm talking to a Portuguese person, it would be easier to use the Portuguese word for book. Likewise, for problems involving rotation, it is more convenient to switch to polar coordinates. If the problem had involved some sort of displacement, Cartesian would be more advantageous. So let's switch to PCS first:

Transformation of P to PCS

In the given problem, I know that the coordinates of P are (x_P, y_P). For simplicity, I will take them as (x, y) and do the transformation. r is the distance here:

Since P0x is a right triangle, simple trigonometric equations for θ can be easily written. E.g. sin(θ) = y/r and cos(θ) = x/r . If I divide these two side by side tan(θ) = y/x (r cannot be zero, btw. If it was, P would already be on the origin and the angle would be undefined, since there is no line segment, no triangle. Therefore the division is valid). Hence:

The equations for the inverse transformation are shown in the image below.

Equation (1)

Now, we can proceed to the solutions.

Solution

a) Rotating the coordinate system S by φ means, that we're expressing the point P(x’, y’) as P(r, θ - φ) in the new coordinate system. In this case, the following equations are written from the sum-difference identities for trigonometric functions:

If I open the parentheses on the right hand side, and substitute the terms by x and y from the equation (1), the following equations are found:

If I express this with a matrix:

Equation (2)

 

b)  

To show the relationship between x' and x, as well as y' and y, equations (1) will be used again. Since x and y are real numbers, the expressions in the roots can never be negative. Therefore, by squaring both sides, we can get rid of the roots without doing any mistake. The equation is shown below:

 

In the above equation, I assumed that the equation  cos²(x) + sin²(x) = 1 is already known. Considering the right triangle P0x in the image where P is shown in CCS, the hypotenuse is r and other edge lengths are r*cos(θ) and r*sin(θ), which are obviously perpendicular to each other. The equality is easily proved by writing down the Pythagorean theorem and simplifying all terms by  r². The geometric proof of the equality for the case r = 1 can also be easily given in an unit circle.

c)

We now want to show the above equality. Once again, the equations in (1) will be used here. The expressions inside the roots are not negative, so we get rid of the roots by squaring both sides. From the rotation transformation given in (2), the following equations are written:

As these are relatively complicated, I first opened the terms with x, and then the terms with y.

 

 

To get the expression (x'_P - x'_Q)² + (y'_P - y'_Q)² , I will add the above terms side by side. The terms marked in same color above cancel each other out. For the sake of simplicity, they are excluded in the summation below.

Finally, if the right hand side of the above equation is grouped by x's and y's, and bracketed, and finally the resulting equation is simplified based on  cos²(x) + sin²(x) = 1 , a square bracket expansion is found:

[1]: Moebs, W., Sanny, J., & Ling, S. J. (2016). University Physics Volume 1. Houston, Texas: OpenStax. Link