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.

  • $\begingroup$ You forgot a square $|\bar a \times \bar b|^\color{red}{2}$ on the LHS of 1.. Both identities follow fairly straight from the definitions of the vector products. What have you tried and where did you get stuck? $\endgroup$ – dxiv Apr 8 '17 at 1:57
  • $\begingroup$ It appears you are using \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
  • $\begingroup$ @dxiv It is true. I have forgot the exponent $^\color{red}{2}$. With so much sincerity I simply noticed that the formula 1. looks like a binomial squared, but being $|\vec{a}\times\vec{b}|^2=|ab\sin \vartheta|^2=|a^2b^2(1-cos^2\vartheta)|$ i'm not sure how to continue. $\endgroup$ – user401938 Apr 8 '17 at 7:24
  • $\begingroup$ @hardmath You're certainly right, now I add the details. Yesterday I could not do it. $\endgroup$ – user401938 Apr 8 '17 at 7:29
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    $\begingroup$ @Sebastiano, Have you tried the hint from tommy xu3 below? The proof using his hint should be much simpler than what you did. $\endgroup$ – Hoc Ngo Apr 14 '17 at 20:50

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: $$\begin{align} |\vec a + \vec b|^2 &= (\vec a + \vec b )\cdot(\vec a + \vec b ) \\ &= \vec a \cdot\vec a + \vec a \cdot \vec b + \vec b \cdot \vec a + \vec b \cdot \vec b\\ &= a^2 + b^2 + 2 \vec a \cdot \vec b. \end{align} $$



$1.$ $|\vec{a}\times\vec{b}|=|ab\sin\theta|$ and $|\vec{a}\cdot\vec{b}|=|ab\cos\theta|.$

$2.$ $|\vec{a}+\vec{b}|^2=( \vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}).$


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