Why doesn't the dot product give you the coefficients of the linear combination? So the setting is $\Bbb R^{2}$.
Let's pick two unit vectors that are linearly independent.  Say: $v_{1}= \begin{bmatrix}  \frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{bmatrix}$ and $v_{2} = \begin{bmatrix}  \frac{\sqrt{3}}{2} \\ \frac{1}{2}\end{bmatrix}$.
Now, let's pick another vector with length smaller than $1$, say, $a = \begin{bmatrix}  \frac{1}{2} \\ 0\end{bmatrix}$.
I've been trying to understand the dot product geometrically, and what I've read online has led me to believe that $a \cdot v_{1}$ is the scalar $c$ so that $cv_{1}$ is the "shadow" of $a$ on $v_{1}$.  Similarly, $a \cdot v_{2}$ is the scalar $d$ so that $dv_{2}$ is the "shadow" of $a$ on $v_{2}$.
If this is true, then it should be that $cv_{1} + dv_{2} = a$, right?  But this isn't the case.
We have $a \cdot v_{1} = \frac{1}{4}$ and $a \cdot v_{2} = \frac{\sqrt{3}}{4}$.  So $$cv_{1} + d v_{2} = \frac{1}{4}\begin{bmatrix}  \frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{bmatrix} + \frac{\sqrt{3}}{4}\begin{bmatrix}  \frac{\sqrt{3}}{2} \\ \frac{1}{2}\end{bmatrix} =  \begin{bmatrix}  \frac{1}{2} \\ \frac{\sqrt{3}}{4}\end{bmatrix} \neq a.$$
This means something is wrong with my understanding about the intuition of the dot product.  I'm not sure what's wrong with it, though.  Any help would be appreciated.
 A: Your intuition is mostly correct, and you would probably have seen the flaws in your reasoning if you had drawn a picture like this:

We have two linearly-independent unit vectors $\mathbf{U}$ and $\mathbf{V}$, and a third vector $\mathbf{W}$ (the green one). We want to write $\mathbf{W}$ as a linear combination of $\mathbf{U}$ and $\mathbf{V}$. The picture shows the projections $(\mathbf{W} \cdot \mathbf{U})\mathbf{U}$ (in red) and $(\mathbf{W} \cdot \mathbf{V})\mathbf{V}$ (in blue). These are the things you call "shadows", and that's a good name. As you can see, when you add them together using the parallelogram rule, you get the black vector, which is obviously not equal to the original vector $\mathbf{W}$. In other words
$$
\mathbf{W} \ne (\mathbf{W} \cdot \mathbf{U})\mathbf{U} + (\mathbf{W} \cdot \mathbf{V})\mathbf{V}
$$
You certainly can write $\mathbf{W}$ in the form 
$\mathbf{W} = \alpha\mathbf{U} + \beta\mathbf{V}$, but $\alpha = \mathbf{W} \cdot \mathbf{U}$ and $\beta = \mathbf{W} \cdot \mathbf{V}$ are not the correct  coefficients unless $\mathbf{U}$ and $\mathbf{V}$ are orthogonal. And you can even calculate the coefficients $\alpha$ and $\beta$ using dot products, as you expected. It turns out that 
$$
\mathbf{W} = (\mathbf{W} \cdot \bar{\mathbf{U}})\mathbf{U} + (\mathbf{W} \cdot \bar{\mathbf{V}})\mathbf{V}
$$
where $(\bar{\mathbf{U}}, \bar{\mathbf{V}})$ is the so-called dual basis of $(\mathbf{U}, \mathbf{V})$. You can learn more here.
A: What you are thinking is correct in terms of orthonormal basis:
Suppose that  $\{(e_i):i\in I\}$ is  an orthonormal basis of any vector space $V$  then any vector $x$ can be expressed as $x=\sum _{i=1}^n c_ie_i$
In order to get the $c_j;j=1,2,...n$ we can use the fact that :
$ \langle x,e_j\rangle =\langle \sum _{i=1}^n c_ie_i,e_j\rangle =\sum _{i=1}^n c_i\langle e_i,e_j\rangle=c_j;j=1,2,...,n$
A: Your $v_1$ and $v_2$ need to be orthonormal.  To expand on learnmore's answer, essentially, the reason you need orthogonality for this to work is that if your $v_1$ and $v_2$ are not orthogonal, then they will have a non-zero dot-product $v_1\cdot v_2$.  This means that $v_2$ carries some weight "in the direction" $v_1$.  Your intuition that $c = a\cdot v_1$ is the "amount of $a$ in the direction $v_1$" is correct - keep that intuition!  Similarly, $d=a\cdot v_2$ is the amount of $a$ in the direction of $v_2$.  
However - since $v_1$ and $v_2$ are not perpendicular, the number $c$ has "piece" of $v_2$ in it, and the number $d$ has a "piece" of $v_1$ in it.  So, when you try to expand $a$ in the basis $\{v_1,v_2\}$, you would need an extra term to compensate for the "non-orthogonal mixing" between $v_1$ and $v_2$.
The technical details are as follows.  Since $v_1$ and $v_2$ are linearly independent, we can write 
$$
a = \alpha v_1+\beta v_2
$$  for some scalars $\alpha, \beta$.  Now, take the dot product of $a$ with $v_1$ and expand it out: 
$$
a\cdot v_1 = (\alpha v_1+\beta v_2)\cdot v_1 = \alpha v_1\cdot v_1 + \beta v_1\cdot v_2 = \alpha + \beta v_1\cdot v_2
$$ similarly, expand out $a\cdot v_2$: 
$$
a\cdot v_2 = \alpha v_1\cdot v_2 + \beta 
$$
Those extra terms ($\beta v_1\cdot v_2$ and $\alpha v_1\cdot v_2$) express the non-orthogonality.  Written another way, we have 
$$
\alpha = a\cdot v_1 - \beta v_1\cdot v_2
$$ and 
$$
\beta = a\cdot v_2 - \alpha v_1\cdot v_2
$$ which shows clearly that the correct expansion coefficients have $a\cdot v_j$, but also another piece compensating for the non-orthogonality.  I could go on - you can use matrices and such, but hopefully this is enough to convince you.
