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Prove that for any two vectors $\mathbf a$ and $\mathbf b$, $\lvert \mathbf a \times \mathbf b \rvert^2 + (\mathbf a \cdot \mathbf b)^2 = \lvert \mathbf a \rvert^2 \, \lvert \mathbf b \rvert^2$.

Can someone offer me advice on how to prove this in an easier way? So far, I'm solving it in a really complicated way, by labelling $\mathbf a$ as $(x,y,z)$ and $\mathbf b$ as $(a,b,c)$, then multiplying them out.

So, for the $\lvert \mathbf a \times \mathbf b \lvert^2$ term, I found $(yc-bz)^2 + (za-xc)^2 + (xb-ya)^2$, and then $(\mathbf a \cdot \mathbf b)^2=(ax+yb+zc)^2$.

Is there an easier way?

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3  
This is a nice generalisation. – πr8 Jan 7 at 19:50
up vote 16 down vote accepted

Yep, there is an easier way. Hint:

$$|a \times b| = |a||b|\sin(\theta)$$

and

$$a \cdot b = |a||b|\cos(\theta)$$

where $\theta$ is the angle between the vectors $a$ and $b$.

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I prefer this purely vector method (using the Einstein summation convention) to the trig based method in the answer by Eli Rose. $$ \begin{align*} &\left| a\times b\right|^2+\left(a\cdot b\right)^2\\ =\;&\varepsilon_{ijk}a_jb_k\varepsilon_{ilm}a_lb_m+a_jb_ja_kb_k\\ =\;&a_jb_ka_lb_m(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})+a_jb_ja_kb_k\\ =\;&a_ja_jb_kb_k-a_jb_ka_kb_j+a_jb_ja_kb_k\\ =\;&a_ja_jb_kb_k=\left|a\right|^2\left|b\right|^2 \end{align*} $$ (Where $\varepsilon$ is the Levi-Civita symbol, $\delta$ is the Kronecker delta, and I've used the relation $\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$)

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1  
Cool -- is this Einstein summation notation? What are $\varepsilon$ and $\delta$? – Eli Rose Jan 8 at 0:21
1  
@EliRose Yes it is. I love the summation convention so much that I forget other people don't know about it. :-) I added some explanatory links to the post. – Oscar Cunningham Jan 8 at 0:35

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