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.
\bar
to indicate vector quantities (but\vec
would produce $\vec a,\vec b$), where the undecorated symbols $a,b$ denote scalars (presumably the lengths of the corresponding vectors?). It is a defensible choice of notation but deserves a few words of explanation. $\endgroup$ – hardmath Apr 8 '17 at 5:17