Identity between two vectors I carried out the proofs of these two identities.
If $\vec{a}$ and $\vec{b}$ are two vectors of $\mathbb{R}^2$, therefore
1. $$|\vec{a}\times \vec{b}|^2\stackrel{id}{=}a^2b^2-(\vec{a}\cdot \vec{b})^2$$
 2. $$|\vec{a}+\vec{b}|^2\stackrel{id}{=}a^2+b^2+2(\vec{a}\cdot \vec{b})$$
First proof.
$$|\vec{a}\times \vec{b}|^2=(|\vec{a}\times \vec{b}|)^2=(ab\sin \vartheta)^2=a^2b^2\sin^2\vartheta=a^2b^2(1-\cos^2\vartheta)=a^2b^2-a^2b^2\cos^2\vartheta=$$
Being $\cos\vartheta=\dfrac{\vec{a}\cdot \vec{b}}{ab} \longrightarrow \cos^2\vartheta=\dfrac{(\vec{a}\cdot \vec{b})^2}{a^2b^2}$, I obtain:
$$=a^2b^2-a^2b^2\frac{(\vec{a}\cdot \vec{b})^2}{a^2b^2}=a^2b^2-(\vec{a}\cdot \vec{b})^2
$$
Second proof.
$$|\vec{a}+ \vec{b}|^2=((a_x+b_x)\hat x +(a_y+b_y) \hat y)^2=(a_x+b_x)^2(\hat x)^2+(a_y+b_y)^2(\hat y)^2=$$
Being $|\hat x|^2=|\hat y|^2=(\hat x)^2=(\hat y)^2=1$, I obtain:
$$=(a_x+b_x)^2+(a_y+b_y)^2=a^2_{x}+2a_{x}b_{x}+b^2_{x}+a^2_{y}+2a_{y}b_{y}+b^2_{y}=$$
$$=(a^2_{x}+a^2_{y})+(b^2_{x}+b^2_{y})+2(a_{x}b_{x}+a_{y}b_{y})=$$
$$=a^2+b^2+2(\vec{a}\cdot \vec{b})$$
Being $\vec{a}=a_x\hat x+a_y\hat y,\,\vec{b}=b_x\hat x+b_y\hat y,\,\,|\vec a|^2=a^2,\,\,|\vec b|^2=b^2,\,\, \vec{a}\cdot \vec{b}=a_xb_x+a_yb_y.\,\,_{\square}$
Are there different proofs from what I've shown? I did not have, at least until now, an answer to my question.
 A: In response to Sebastiano's request, I will provide a detailed proof using Tommy xu3's hint. 
First, note that $|\vec V|$ is scalar so we can use any scalar operations on $|\vec V|$.
a) The proof of the first identity is similar to Sebastiano's, but care must be taken to not divide by $ab$ in case the $ab = 0$. Thus, the identity $\sin^2\theta = 1-\cos^2\theta$ should be used instead.
$$\begin{align}
|\vec a \times \vec b|^2 &= (|\vec a \times \vec b|) (|\vec a \times \vec b|) 
\text{, noting scalar multiplication: (...) times (...)}\\
&= (ab\sin\theta) (ab \sin\theta) \\
&= a^2b^2\sin^2\theta \\
&= a^2b^2(1- \cos^2\theta)\\
&= a^2b^2 - (ab\cos\theta)^2 \\
&= a^2b^2 - (\vec a\cdot\vec b)^2. 
\end{align}
$$
The second identity uses hint #2, which comes from:
$$\vec V\cdot \vec V = V^2 \cos 0 = V^2 = |\vec V|^2. $$
Reading the identity above right to left, letting  $\vec V = \vec a + \vec b$, we have:
$$\begin{align}
|\vec a + \vec b|^2 &= (\vec a + \vec b )\cdot(\vec a + \vec b ) \\
&= \vec a \cdot\vec a + \vec a \cdot \vec b + \vec b \cdot \vec a + \vec b \cdot \vec b\\
&= a^2 + b^2 + 2 \vec a \cdot \vec b.
\end{align}
$$
A: Hint:
$1.$ $|\vec{a}\times\vec{b}|=|ab\sin\theta|$ and $|\vec{a}\cdot\vec{b}|=|ab\cos\theta|.$
$2.$ $|\vec{a}+\vec{b}|^2=( \vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}).$
