In my physics II class our professor has us go through through three steps to find the unit vector $\hat r$:
Write the vector $\vec r$
$$\vec r=5\hat i + 5\hat j$$
Find the distance $r$ using Pythagorean's Theorum
$$r=\sqrt{5^2 + 5^2}\approx7.1$$
Then divide each component of $\vec r$ by $r$.
$$\hat r=\frac 5 {7.1}\hat i+\frac 5 {7.1}\hat j\approx.70\hat i+.70\hat j$$
Then we'll solve the equation to find the magnitude and multiply it by $\hat r$
For example the formulas look like $$\vec E= \frac {Kq} {r^2} \hat r$$
From what I understand this is just writing the vector in x and y components, so couldn't $\hat r$ be written as $\hat r=\cos x \hat i+\sin x \hat j$?
Best Answer
Sure, you can write your vector in polar form, and since $\hat r$ is by definition a unit vector, it will have the form you mentioned:
$$\hat r = \cos\theta\hat i + \sin\theta\hat j$$
for some angle $\theta$. Whether this is better or worse than the rectangular form $\hat r = a\hat i + b\hat j$ is a question of context.
I wonder why you are dividing by $r$ in your equation for $\vec E$ since the magnitude has already been factored out of $\vec r$, but perhaps there is a good reason for this; I don't know the context.