stratsticks57jl

2022-07-20

I want to prove
$|\stackrel{\to }{r}-\stackrel{\to }{s}{|}^{3}=\left({r}^{2}+{s}^{2}-2\cdot r\cdot s\cdot \mathrm{cos}\left(\theta \right){\right)}^{3/2}$
where $r=|\stackrel{\to }{r}|$. but I have problems because
$\left({r}^{2}+{s}^{2}-2\cdot r\cdot s\cdot \mathrm{cos}\left(\theta \right){\right)}^{3/2}=\left({r}^{2}+{s}^{2}-\stackrel{\to }{r}\cdot \stackrel{\to }{s}{\right)}^{3/2}=\left(\left(\stackrel{\to }{r}-\stackrel{\to }{s}{\right)}^{2}{\right)}^{3/2}=\left(\stackrel{\to }{r}-\stackrel{\to }{s}{\right)}^{3}$
I hope someone can help me.

bardalhg

Expert

This is a standard derivation. Start with the fact that $|\stackrel{\to }{v}{|}^{2}=\stackrel{\to }{v}\cdot \stackrel{\to }{v}$ applied to $\stackrel{\to }{v}=\stackrel{\to }{r}-\stackrel{\to }{s}$ to get
$|\stackrel{\to }{r}-\stackrel{\to }{s}{|}^{2}=\left(\stackrel{\to }{r}-\stackrel{\to }{s}\right)\cdot \left(\stackrel{\to }{r}-\stackrel{\to }{s}\right).$
Now expand the dot product to get
$|\stackrel{\to }{r}-\stackrel{\to }{s}{|}^{2}={r}^{2}+{s}^{2}-2\stackrel{\to }{r}\cdot \stackrel{\to }{s}={r}^{2}+{s}^{2}-2rs\mathrm{cos}\left(\theta \right).$
Now take the square root of both sides, then cube. Done.

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