How is the dotproduct a projection? I don't understand who projects on who. How is the following a projection on anything? I don't see how a vector of lenght 6 can possibly result of a projection here. If I project C down to the x-axis, it lands on "2". If I project B upwards towards C it will result in something along "3". Why would it be "6" ?!

 A: The dot product $a\cdot b$ tells you something about the length of the projection of $a$ onto $b$, (or $b$ onto $a$, since the relationship is symmetric.)
That is, if you decompose $a=a_\perp+a_\parallel$ where $a_\perp$ is perpendicular to $b$ and $a_\parallel$ is parallel to $b$, then $a\cdot b=|a_\parallel|$. 
Edit: Actually it will not be exactly that. A corrected version of what I was thinking of follows. Again, decompose $a=a_\perp+a_\parallel$ where $a_\perp$ is perpendicular to $b$ and $a_\parallel$ is parallel to $b$. If $b$ has unit length, then $a\cdot b$ will give the length of $a_\parallel$. If $b$ doesn't have unit length, then the dot product reports $|b||a_\parallel|$ or $-|b||a_\parallel|$, depending on if the angle between $a$ and $b$ is acute or obtuse, respectively.
I would also have to modify this explanation to account for situations where there the dot product is negative by referring to a "signed length". At any rate, I think this is the way to link projection and the dot product.
So, for example, if $a$ and $b$ are perpendicular, then the projection of $a$ onto $b$ is a single point (i.e. a length zero vector) and so $a\cdot b=0$ in that case.
In your example, $a\cdot b=6$, which is, depending on how you look at it, the length of $b$ times the length of $a_\parallel$, ($3\cdot 2$) or the length of $a$ times the length of $b_\parallel$ ($2\sqrt 2 \cdot \frac{3}{\sqrt{2}}$).
