Vectors in a Cartesian coordinate system - homework help I have a homework question to ask that concerns vectors in a Cartesian coordinate frame. I'm not really good with vectors so that question is a bit difficult for me to understand.
Anyway, this is the question:

The figure 1 shows the vector r = ai + bj, in Cartesian coordinate frame. Find the expressions for $e_r$ (unit vector in the direction of r) and $e_\phi$ (unit vector in the direction perpendicular to r). Calculate $e_r$, $e_\phi$ for r = 1i+2j and sketch on Cartesian xy frame.

I have so far solved the first part of the question. Hence, I have the two expressions for $e_r$ and $e_\phi$, which should be:
$$e_r = \cos(\phi)i + \sin(\phi)j$$
$$e_\phi = - \sin(\phi)i + \cos(\phi)j$$
I hope that is correct? I don't know how else should I write it down since the question requires expression and not e.g. components of $e_r$ and $e_\phi$.
Then I have some issues with the 2nd part of the question concerning the calculation of $e_r$ and $e_\phi$ with known values of a and b. I have been trying to solve the question with the help of polar coordinates rules and derivation of $e_r$ and $e_\phi$ with r but with no success. Now I was thinking to work on the solution with the help of unit vectors but I'm not sure. How should I solve it? Is there any specific way?
I don't want an answer to the homework question but I'd really appreciate an explanation on how to reach the solution of this question or rather what's the process to be able to understand such tasks in the future. I'd really appreciate all the help!
 A: Ok, there are two ways to think about vectors. I will give the intuitive "natural" way and then the mathematically precise way.
Intuitively vectors are simply "things"(whatever you want them to be) that can have both a magnitude and a direction.
Temperature is not a vector because it only has magnitude(or you could call it direction, but it only has one, not both).
A plane moving through the air has a vector. This means it has both a direction and a magnitude. The magnitude here is the air speed(or we could use altitude, or any parameter, actually).
Mathematically, a vector is an element of a Vector space(a space of vectors, a set of vector with additional properties).
If V, a set of elements[called vectors], and a field K(e.g., your typical real numbers) obey the following or have the following properties then V can be called a Vector Space or has the properties of a vector space. The following is simply a definition of a Vector Space:
V forms an additive abelian group:

*

*0                    - additive identity

*a+(b+c) = (a+b)+c    - associativity

*a+(-a) = 0           - additive inverses

*a+b in V             - closure

*a+b = b+a            - commutativity

V over K forms a distributive group

*

*k*(a+b) = ka + kb  - K distributes on the left of V

*(a+b)k = av + b*v  - K distributes on the right of V

*v*(k+s) = vk + vs  - V distributes on the right of K

*(k+s)v = kv + s*v  - V distributes on the left of K

*(kl)v = k(l*v)     - V is compatible with K

*k*v \in V            - K is compatible with V

*1*v = v              - K is idempotent on V

The mathematically precise definition may seem extremely complex but, in fact, it is quite simple. Y ou have to just read through it and understand what the terms mean. Y ou already know all this stuff since you have been using them your entire life. You just don't realize they have specific names because they are very helpful concepts. E.g., 3 + 5 = 5 + 3 which expresses commutativity. The reason why you know you can add in either order is because you know + is commutative. You actually know it works this way but you don't necessarily understand the naming scheme.
R^2 over R is a vector space. The reason is because it was designed this way. A vector space just comes from taking how we add on single numbers and turns them in to adding on multiple numbers: E.g., (3,8) + (5,4) = (8,12).
Once you do this then you can start doing math on entire vectors of numbers rather than one at a time.
E.g., suppose you want to add 2 + 4 and 5 + 8  then multiply by 3? You can first do 2 + 4 = 6 then 36 = 18, and then 5 + 8 = 13 then 313 = 39. Or you can do 3*((2,5) + (4,8)) = 3*(6,13) = (18,39).
So vector spaces are just "extensions" of the concept of fields.
The ultimate idea to understand is that they are just familiar numbers, we are just grouping them in a natural way so they mimic additive real numbers that we learned in first grade.
But in math we don't want the familiar, we use it only as a bridge in to the unknown. So at some point you have to start thinking of a vector space as it's own entity. It is more general, but not entirely different, from a R^n which you come to know and love.
That is, you think of vectors as more abstract things that you don't even know what they represent or contain but you do know they are vectors and hence have the properties of vectors. It's then becomes algebra. One day then you can do calculus on the vectors. At the end of the day, a vector is very simple, someone pointing is a vector. They are expressing the concept of a direction and distance('It's that far over there' = some vector).
Vector spaces are higher levels of though about these concepts so learning to think of them will make you more intelligent(or possibly less depending on your orientation).

Since you are talking about Cartesian coordinates you are talking about R^2 over R = Euclidean 2-space. They are isomorphic structures.
Now, what the question is asking you to do is convert from one basis of R^2 to another basic of R^2. First from the standard then to the cylindrical.
to do this you must recall the formulas or logic(hence knowing how formulas are arrived at is very important as it is what provides the continuity of being able to recall. Do you know how one gets the transformation formulas that are used in this problem? If not you will have to derive it all from scratch which might be very difficult. This is why we step through and learn proofs so we have a foundation to apply the results of the theorem. Without understanding the proofs, even though they seem hard at first they get easier the more one works at it, one won't really know what is going on. The proofs are the glue that tie all the definitions together and provides the structure and meaning that the theorem brings.
The standard coordinate basis are the simplest. The reason is is that they are constant coordinate functions. This gets a little in to manifolds and curvlinear coordinates but just know that the standard coordinate basis of a vector space is the simplest one can choose to represent a system. This "simplest" is simple in that we can write out vectors in single dimensions of the field K: v = (k1,k2,k3...,kn). R^n over R is just a vector that we can write as (a,b,c,...,n) with a,b,c,...n \in R.
Now cylindrical coordinates are curvlinear but really they are just a different basis. Luckily we can translate from one to the other and this is the simple "$xy$" to "$r\theta$" conversion:
$r = \sqrt{x^2 + y^2}$ and $\theta = tan^{-1}\left(\frac{y}{x}\right)$
and we can go back(they are topologically equivalent):
$x = r\cos(\theta)$ and $y = r\sin(\theta)$.
The idea of a basis is that it gives us a way to represent, or break down, a vector in to components of this basis. The basis then acts as building blocks for making combinations to create new vectors.
Different basis will present the space or any sub-space in it's own character. Generally the idea is to find the best basis representation as that is truly the most simple as far as the inherent nature of the problem/data is concerned. E.g., if something is cylindrical then thinking about it in terms of cylindrical coordinates will give the most simplest way to discuss the problem. E.g., with Cartesian coordinates it requires much more difficulty to understand a cylinder.
Those equations I gave above can be simplified as vector transformations though. Since it is the transformations that matter, not the actual values(they are just place holders more or less), we can create a conversion of $(x,y)$ to $(r,\theta)$:
$(x,y) \rightarrow \left(\sqrt{x^2 + y^2}, \tan^{-1}\left(\frac{y}{x}\right)\right)$
Or the same as:
$(r,\theta) = \left(\sqrt{x^2 + y^2}, \tan^{-1}\left(\frac{y}{x}\right)\right)$
Remember that $(a,b)$ is "Cartesian-coordinate point notation" while $a e_x + b e_y$ "vector component representation" but they are just the same. Also
Remember that $(a,b)$ is "cylindrical-coordinate point notation" while $a e_r + b e_\theta$ "vector component representation" but they are just the same. Also ...
So, unfortunately $(a,b)$ can be ambiguous but usually it is for Cartesian coordinates.
So your problem is quite simple: You must pattern match then simplify(It's the same pattern for simplification processes). You must recognize the components and which space they are in(usually given by the problem) and then you must translate them to the space you want them in(the space is usually told to you in the presentation of the problem). The translation part is finding the right formulas and doing the computation.
It is a 3 part problem with really only the middle part being any work.
It feels like you have several weaknesses: You are not that great and understanding geometry. You have vague idea but just overall weak on the subject. You you are weak on algebra. You also do not have a lot of experience with physics or mechanics. While as you study you will learn this stuff you might want to focus on your past weaknesses so they don't slow you down any more. Everything in math is built up from the floor and any weakness in the foundation will become a weakness transmitted in to all future layers. This is not to bad though as you will learn and those weaknesses can become strengths as long as they do not stop you from learning.
Try to learn to think of these mathematical things in terms of real world concepts, they almost always apply to something. If you can visualize something based on them then it will help to understand what is really going on.
In this case we are just trying to convert something in xy coordinates to something in circular coordinates. This is the most basic idea you should get from the problem. I've just explained it in more detail and a way that helps you understand that it works in general. You want to learn the general way so you can convert in to spherical coordinates, sympletic coordinates, or any arbitrary curvlinear coordinates which will then give you a way to really understand the universe in far more complexity.
The concept of a function is very important. A function takes something and changes it according to "the function" and then gives that thing back.
You are a function. You do something, you change it, and then "give it back". A mathematical function takes an object and changes it and gives it back.
A transformation is just a mathematical function that usually entails some type of regularity in the function(this is why we call it a transformation rather than just a function. A transformation is a little more than a function but still a function. Just like spaghetti is still noodles but it's more than just noodles).
I also think you fail to understand precisely what is the difference between coordinates and basis.
A basis is simply what we use to spell out coordinates. $(a,b,c,d)$ represents 4 coordinates and the command simply separate the coordinates. This lets us keep things nice and organized. Remember, all this stuff comes from just working with individual elements but we combine them in to being able to do work in parallel(parallel is faster once you get the hang of it). You could call that the comma basis but really it's just a way to say $a\hat{i} + b\hat{j} + c\hat{k} + d\hat{l}$ where the $\hat{\;}$ sorta acts like a comma. They serve the purpose to separate and divide up.
BUT since our data(elements) generally come from some specific representation(e.g., in the physical world they will come from some measurement devices) we have to know that. Don't think of data as just numbers, they are numbers but also came out of something. Even if a calculator spits out 42, an entire process happened before that. The 42 has attached an entire evolution to it.
So $(3,4)$ doesn't mean anything except relative to the "standard" basis(which has no hidden meaning). We can write it as $3\hat{i} + 4\hat{j}$. The $\hat{i}$ is sort of "first coordinate" and the $\hat{j}$ is sort of "second coordinate"
But if really that $(3,4)$ is representing radians and ft/sec then we might, if we can do it, translate those in to "standard coordinates" which might give us $(5.43, 0.2443)$ which we might be able to right as $5.43\hat{\theta} + 0.2443\hat{s}$.
$(3,5)$ is like giving the first name, you don't know the other part of the equation. Someone somewhere has to tell you where those 3 and 5 came from or what they represent, else you can't know. You can assume it is standard basis and get away with doing all your math, then if they tell you "Opps, that was suppose to be in cylindrical coordinates" then guess what? Absolutely nothing is wrong because, as long as you were consistent with only using the vector space properties you will then be able to convert to cylindrical coordinates at the end and arrive at the correct solution.
e.g., $(3,5) + (4,2) = (7,7)$ Now if it were to convert to cylindrical coordinates: $C[(3,5) + (4,2)] = C[(7,7)]$ where here I use $C[(a,b)]$ to represent "convert the point $(a,b)$ to cylindrical(or whatever) coordinates.
If C is "linear" then this means $C[(3,5) + (4,2)] = C[(3,5)] + C([4,2])$.
What this says is that, for linear transformations at least, we can either convert first or convert last. Either way we will get the same answer(this is commutativity, linear transformations commute with + in a vector space).
Linear transformations are what one starts to get in to with vector spaces because they are quite powerful
A: The given vector is
$ v = (1, 2) $
Hence,
$e_r = \dfrac{v}{|v|} = \dfrac{(1, 2)}{\sqrt{1^2+2^2}} =\dfrac{(1, 2)}{\sqrt{5}} = (\dfrac{1}{\sqrt{5}}, \dfrac{2}{\sqrt{5}})$
Note that if $e_r = (x, y)$, then $e_\theta = (-y, x) $
Therefore, in our case,
$e_\theta = (-\dfrac{2}{\sqrt{5}}, \dfrac{1}{\sqrt{5}} )$
