TITLE: Why doesn't the dot product give you the coefficients of the linear combination?
QUESTION [13 upvotes]: 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.

REPLY [8 votes]: 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.