How $|\vec{a} +\vec{b}|^2 = |\vec{a}|^2 +|\vec{b}|^2 + 2\vec{a}\cdot \vec{b}$? Please help in to understand why/how $$|\vec{a} +\vec{b}|^2 = |\vec{a}|^2 +|\vec{b}|^2 + 2 \vec{a}\cdot \vec{b}$$
My precise confusion is why the modulus is not included in $2 \vec{a}\cdot \vec{b}$ ?
Since if we do 
$(|\vec{a}| +|\vec{b}|)^2$, we have modulus sign in every $\vec{a}$ and $\vec{b}$.

Let us write the norm in terms of inner product, namely $|x| = \sqrt{\langle x,x\rangle}$.
Now: 
$$\begin{align*}
|a+b|^2 &=\langle a+b,a+b\rangle\\
 &= \langle a,a+b\rangle + \langle b,a+b\rangle \\
 &= \langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle +\langle b,b\rangle\\
 &= |a|^2 + |b|^2 + \langle a,b\rangle +\overline{\langle a,b\rangle}\\
 &= |a|^2 + |b|^2 + 2Re(\langle a,b\rangle)
\end{align*}$$
Where $Re$ is taking the real part of a complex number, and of course when the vector space is real then $2Re(\langle a,b\rangle) = 2\langle a,b\rangle$.
