Covariant Contravariant Dot product and Length So one thing that I find really interesting is that if I have a vector $\vec V = V_x \hat i + V_y \hat j$ its length is just:
$$ |V|^2 = V_x^2 + V_y^2 $$
That is all well and good, but then if I transform the vector into a new basis, I can rewrite the vector in terms of a covariant basis as:
$$ \vec V = V^1 \vec b_1 + V^2 \vec b_2 $$
Now of course, it is obvious that since $\vec b_1 $ and $\vec b_2$ are not necessarily orthogonormal, that $|V| \ne \sqrt{(V^1)^2 + (V^2)^2}$, that is all well and good:
Contravariant Basis
Now the usual way this goes is that we then define a new set of basis vectors: we define $b^1$ to be orthogonal to all $b_i$ when $i\ne 1$ but we define strangely that $b_1 \cdot b^1 = 1$. Then we rinse and repeat for all other vectors.
We can then represent v in terms of this new basis directly as:
$$ \vec V = V_1 \vec b^1 + V_2 \vec b^2 $$
Now this is also fine, but then something totally out of the blue happens:
The Dot Product
If we take the dot product of these two representations, we can get an alternative formula for the length:
$$ |V| ^2 = (V^1 \vec b_1 + V^2 \vec b_2) \cdot (V_1 \vec b^1 + V_2 \vec b^2) = V_x^2 + V_y^2 $$

Now I can verify this by calculation, but I now realise that I have absolutely no understanding of why this should be true. 

Any help would be most appreciated :) I don't see the connection here, why does defining this new basis with the rule that $b_j \cdot b^k = \delta_{jk}$ lead to such an elegant formula for length?
 A: The intuition here is as follows: we define the dual basis to "correct for" all the departures from orthonormality of the original basis. So if the angle between two basis vectors in the original basis was acute, the angle in the new basis will be obtuse; if one if the basis vectors was longer in the original basis, it will be shorter in the new basis. Then when we "average out" the two bases by taking the product of the coordinate in one basis with the coordinate in the other basis, all the departures from orthonormality are corrected for, and we get the appropriate length as if we had a single orthonormal basis.
Of course, to describe precisely how this "correction" is done and to prove that it works, you have to go through the math, as, e.g., the book Giuseppe Negro pointed you to does.
Did you have a particular problem with the proof there? Or were you looking for intuition?
A: Okay, so I suppose its time I come back and answer my own question here. Its wasn't too
complicated, but thats the beauty of learning isn't it! 
I was missing the fact that we define the dual basis as the basis that makes 
$\bf{e_i} \cdot \bf{e^j} = \delta_i^j$. Given this fact alone, it is possible to find
all of the dual basis vectors $\left\{\bf{e^j}\right\}$. So using my old notation,
(but bolding instead of adding all those ugly hats), we always know that the squared magnitude equals the dot product of any vector with itself, so:
$$ |V|^2 = \bf{v} \cdot \bf{v} $$
So we can write $\bf{v}$ in terms of the original basis, and in terms of the dual basis as I've shown below:
$$ |V| ^2 = (V^1 \bf {b_1} + V^2 \bf {b_2}) \cdot (V_1 \bf {b^1} + V_2 \bf{b^2})$$
Now if you expand this out, you'll get:
$$ |V| ^2 = V^1 V_1 \bf {b_1}\cdot\bf {b^1} + V^1 V_2 \bf {b_1}\cdot\bf {b^2} +
V^2 V_1 \bf {b_2}\cdot\bf {b^1} + V^2 V_2 \bf {b_2}\cdot\bf {b^2}$$
Now just by using our definition, we can immediately say that $\bf{b_1}\cdot\bf{b^1} = \bf{b_2}\cdot\bf{b^2} = 1$. We have also stated that the dual basis vectors are perpendicular to all original basis vectors (except the one with the same index), so $\bf{b_1}\cdot\bf{b^2} = \bf{b_2}\cdot\bf{b^1} = 0$. If we make these substitutions, we get the simple formula:

$$ |V| ^2 = V^1 V_1 + V^2 V_2 $$

Which I was struggling to understand in the past.
