**I do not quite get the part of p. 523 regading Boas, 3rd, Eqn. 8.9. How do we know that the vector \vec{a}_\theta has to be divided by a scale factor to get the unit vector \vec{e}_\theta?**

\vec{a}_\theta = -\hat{x} r \sin(\theta) + \hat{y} r \cos(\theta)

\vec{e}_\theta = \frac{\vec{a}_\theta}{|\vec{a}_\theta|} = -\hat{x} \sin(\theta) + \hat{y} \cos(\theta)

A unit vector is defined to be a vector that has a magnitude of 1. The magnitude of \vec{a}_\theta can be calculated by taking the square root of the sum of the squares of its components. This results in the quantity r which is in general not equal to 1. To create the corresponding unit vector \vec{e}_\theta, you just need to divide \vec{a}_\theta by its magnitude which is just r in this case.