Rotation of Vector in P: Why Divide by r?

[Soff]

When you rotate passivly an arbitrary vector in P (which is not in the origin of a coordinate system), you will get the following new coordinates for the same vector:

x'=x \cos\theta+y\sin \theta
y'=\frac{1}{r}(y\cos \theta-x\sin\theta)

where r is the distance from the origin to the point P and x,y are the components of the vector. Can somebody explain me, why you have to divide by r in the second equation?

 

[HallsofIvy]

You don't! For one thing, there is no "r" in the original information- unless you are assuming that r is the length of the vector <x, y>. But, in any case, there is no "r" in the correct formulas:
x&#039;= x cos(\theta)+ y sin(\theta)
y&#039;= y cos(\theta)- x sin(\theta)
which are simply what you give without the "r".

You can check by taking x= 2, y= 0 and rotating through \theta= 90 degrees.
Assuming, again, that r is the length of the vector, then here r= 2 and your formulas give x'= 2(0)+ 0(1)= 0, y'= (1/2)(2(1)- 1(0))= 1 but <0, 1> is a a vector with length 1.

Using the formulas without the r, you get x'= 2(0)+ 0(1)= 0, y'= 2(1)- 1(0)= 2 giving the vector <0, 2> which is correct.

 

[Soff]

I attached the graphic which describs the situation with the rotation. What I want is to describe the vector in the lower part of the system in coordinates of A^{\theta} and A^{r}. The professor told me that the solution is:

A^{r}=A^{x}\cos\theta+A^{y}\sin\theta
A^{\theta}=\frac{1}{r}(A^{y}\cos\theta-A^{x}\sin\theta)

The first component is simply a rotation, but why this factor \frac{1}{r} in the second component?

 

[Doodle Bob]

Why don't you ask the professor?