daniko883y

2022-09-07

$|\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }{|}^{2}={r}^{2}+{r}^{\prime 2}-2r\cdot {r}^{\prime }\mathrm{cos}\left(\theta \right)$
How is this possible?
I could understand why, if we had brackets instead of abs values. But as it is I cannot understand how to operate with the square of an abs. value

Zara Pratt

Since $\stackrel{\to }{r}$ and ${\stackrel{\to }{r}}^{\prime }$ are elements of a vector space, and not elements of an integral domain, $|\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }|$ does not denote absolute value, but the norm (magnitude) of the vector $\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime },$, although it actually would be more correct to denote it as $||\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }||$. By definition, $||v||=\sqrt{v\bullet v},$, where $\bullet$ denotes the inner product (the dot product, in this case). So
${||\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }||}^{2}=\left(\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }\right)\bullet \left(\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }\right)=\stackrel{\to }{r}\bullet \stackrel{\to }{r}-\stackrel{\to }{r}\bullet {\stackrel{\to }{r}}^{\prime }-{\stackrel{\to }{r}}^{\prime }\bullet \stackrel{\to }{r}+{\stackrel{\to }{r}}^{\prime }\bullet {\stackrel{\to }{r}}^{\prime }={||\stackrel{\to }{r}||}^{2}+{||{\stackrel{\to }{r}}^{\prime }||}^{2}-2\stackrel{\to }{r}\bullet {\stackrel{\to }{r}}^{\prime }\mathrm{cos}\left(\theta \right).$
This, is actually, the law of cosines in disguise.

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