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 $\vec{b}$.
 A: 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 $2Re(\langle a,b\rangle) = 2\langle a,b\rangle$.
A: The problem is that $|\vec a+ \vec b|^2 \neq (|\vec a| + |\vec b|)^2$, because $|\vec a + \vec b| \neq |\vec a| + |\vec b|$.
(Suppose $\vec a$ and $\vec b$ both have length 1 but have opposite directions; then you can see why it's not equal!)
You have to use the fact $|\vec a|^2 = \vec a \cdot \vec a$ and properties of the dot product to prove the identity you have above.  Using the fact that the dot product of a vector with itself is the square of the modulus, your identity is the same as $(\vec a + \vec b) \cdot (\vec a + \vec b) = \vec a \cdot \vec a + 2 \vec a \cdot \vec b + \vec b \cdot \vec b$.  Now can you see why it's true?

Note: I use the LaTeX code \cdot to write the dot product.  Even though they have some of the same properties, the dot product is not the same as multiplication.
A: Because...
$|\vec{a} +\vec{b}|^2 = (\vec{a} +\vec{b}) \cdot  (\vec{a} +\vec{b})$
$(|\vec{a}| +|\vec{b}|)^2 = (|\vec{a}| +|\vec{b}|)\cdot (|\vec{a}| +|\vec{b}|)$
