Simple dot product question u and v are vectors
\begin{align*}
\|u+v\|  &= 4 \\
\|u\| &= \|v\| = 3\\
\end{align*}
find $u.v$
$u.u= \|u\|^2 \ = 9 and\ v.v=\|v\|^2 \ and (u+v).(u+v)=u.u+2u.v+v.v $
so $u.v = \frac{(16-9-9)}{  2} = -1$
My question is why is $u.v$ not equal to $9$?
Im sorry I have forgotten the basics. 
 A: $\|u\|=\sqrt{u.u}$. Now,
$$(u+v)\cdot (u+v)=u\cdot v+2u\cdot v+v\cdot v=\|u\|^2+2u\cdot v+\|v\|^2=18+2u\cdot v$$
It follows
$$u\cdot v=\frac{1}{2}[(u+v)\cdot (u+v)-18]=\frac{1}{2}[\|u+v\|^2-18]=\frac{1}{2}(16-18)=-1$$
Norm is not multiplicative: $\|u\cdot v\|\neq \|u\|\cdot\|v\|$, it is only homogeneous: $\|\lambda v\|=|\lambda|\|v\|$, for every scalar $\lambda$
A: Well there's no "intuitive" reason as to why, because at least as far as I know it dot product is just a specific case of an inner product, one can have many different inner products products defined as long as they follow some rules:
Over $\mathbb{R}$ :
An inner product  $(,):V\times V\to\mathbb{R}$:


*

*Linear with respect to the first parameter: $\left(u+u^{'},v\right) = \left(u,v\right) +\left(u^{'},v\right) $.

*Homogeneous with respect to the first parameter: goven $a \in \Bbb{R} $ then $\left(au,v\right) = a\left(u,v\right)$.

*Commutative : $\left(u,v\right) = \left(v,u\right)$

*Positive $u\in V , u\neq 0_{V} \Rightarrow \left(u,u\right) >0 $


A bit different Over $\mathbb{C}$ :
All first three are the same then there's a slight difference:


*

*Conjugate symmetry : $\left(u,v\right) = \overline{\left(v,u\right)} $

*Positive : $u\in V , u\neq 0_{V} \Rightarrow \left(u,u\right) >0 $ which actually contains two properties :

*

*The dot product of a non zero vector with itself is real.

*The dot product of a non zero vector with itself is positive



So you actually can define a lot of different inner product products and what you're specifically talking about is the standard inner product (at least that the way my linear algebra prof calls it :) ) which we commonly use.
As to a specific example: what you did what correct, but consider the vectors :$v = \left(\begin{matrix}\frac{3}{\sqrt{2}}\\
\frac{3}{\sqrt{2}}
\end{matrix}\right) , u = \left(\begin{matrix}- \frac{3}{\sqrt{2}}\\
\frac{3}{\sqrt{2}} 
\end{matrix}\right) $ in $\mathbb{R}^2$ 
You can see that $||u|| = ||v|| = 3 $ but $u\cdot v = 0$  so the most intuitive thing I can propose is to think about the fact that there are many vectors with the same length but that does not imply anything specific on their coordinates , so their dot product can be different then their individual inner product.
