Sorry if this question is quite elementary, I'm not very good at linear algebra. If the dot product of two vectors $\mathbf{a}\cdot\mathbf{b}=c$ then how do I solve for $\mathbf{a}$? I thought I could work it out by turning it into regular algebra ($c = a_1b_1+a_2b_2\ldots$). But after this step I'm stuck already. Please explain the answer if possible.
[Math] How to solve to get vector from a dot product
linear algebraproductsvectors
Related Solutions
Really, there isn't a notation that is more correct. It is just a matter of convention. All of them mean the operation $\sum_{i = 1}^n a_ib_i$. The important thing is that you understand what you must do. Like you said yourself, in $\mathbf{A \cdot B^T}$, we see $\mathbf{A}$ and $\mathbf{B}$ as row vectors. The $\mathbf{^T}$ serves just to remind you that you can see the dot product as a matrix multiplication, after all, we will have a $1 \times n$ matrix times a $n \times 1$, which is well defined, and gives as result a $1 \times 1$ matrix, i.e., a number.
The notation $\mathbf{A \cdot B}$ doesn't sugest any of these things, and you can think directly of the termwise multiplication, then sum.
In Linear Algebra, we often talk about inner products in arbitrary vector spaces, a sort of generalization of the dot product. Given vectors $\mathbf{A}$ and $\mathbf{B}$, a widely used notation is $\langle \mathbf{A}, \mathbf{B} \rangle$. An inner product (in a real vector space), put simply, is a symmetric bilinear form (form means that the result is a number), which is positive definite. That means:
i) $\langle \mathbf{A}, \mathbf{B} \rangle=\langle \mathbf{B}, \mathbf{A} \rangle $;
ii) $\langle \mathbf{A} + \lambda \mathbf{B}, \mathbf{C} \rangle = \langle \mathbf{A}, \mathbf{C} \rangle + \lambda \langle \mathbf{B}, \mathbf{C} \rangle$ ;
iii) $\langle \mathbf{A}, \mathbf{A} \rangle > 0 $ if $\mathbf{A} \neq \mathbf{0}$
I, particularly, don't like the notation $\mathbf{A \cdot B^T}$, because when working in more general spaces than $\Bbb R^n$, we don't always have a finite dimension, so matrices don't work so well. I never saw a notation different from those three I talked about. But I enforce what I said at the beginning: there isn't a correct notation, but you should be used to all of them, as possible.
Write $a \cdot b = ( |a|) ( |b| \cos \theta)$
Then consider 
The projection of B onto A is of length $|b| \cos \theta$, it is the portion of the green line up to the dotted line perpendicular to it (or I suppose beyond it). Then the dot product is the just magnitude of these two vectors, $a$ and $b\cos \theta$, multiplied.
Note: I have interchanged $a$ and $A$, $b$ and $B$ due to the diagram and your question.
Have a look in the preview of this book, Section 1.2 has a good explanation of the dot product:
Best Answer
It should be pretty clear that there are infinitely many solutions $\mathbf a$, which means that we can't "solve for $\mathbf a$".
Let's look at an example in the $2$-D case: take $\mathbf b = (2,3)$, and consider $$ \mathbf a \cdot \mathbf b = 2 \implies 2a_1 + 3a_2 = 2 $$ Notice that this equation is solvable for any value of $a_2$, since we have $$ 2a_1 = 2 - 3a_2 \implies a_1 = 1 - \frac 32 a_2 $$ Which is to say we can consider $a_2$ to be a free variable. So, for example, $\mathbf a = (1,0),(- \frac 12, 1),(-2,2),(-5,4)$ are all solutions to this equation.