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Consider two unit vectors $u, v$ and name the angle between them as $\theta$. My claim is that \[ \lim_{\theta \to 0} \frac{\theta}{|u - v|} = 1. \]

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I made the title more descriptive and changed the statement to something involving a limit, which seems to make more sense. On the other hand, if you don't know much about limits and just want a heuristic (some people here are experts with pictures) then please say so! – Dylan Moreland Jan 13 '12 at 16:34
up vote 1 down vote accepted

Yes, your claim is correct....

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Umm, huh? If $\theta = 90^\circ$ then $|u-v| = \sqrt2$ but $\sin \theta = 1$? – Scaramouche Jan 13 '12 at 10:35
You forgot that $\theta \rightarrow 0$. – user17090 Jan 13 '12 at 10:37
My point is, how do you figure that $|u-v|=|u \times v|$? – Scaramouche Jan 13 '12 at 10:43
Sorry, I wanted to delete my post but I couldn't. – user17090 Jan 13 '12 at 19:19

Let us denote : $|u-v| = p$ . According to the Cosine Law we can write :

$p^2=|u|^2+|v|^2-2\cdot |u|\cdot|v|\cdot \cos \theta \Rightarrow$

$\Rightarrow p^2=2 \cdot(1- \cos \theta)=2\cdot 2 \cdot \sin^2 {\frac{\theta}{2}}=4 \cdot \sin^2 {\frac{\theta}{2}} \Rightarrow$

$ \Rightarrow p=2\cdot \sin {\frac{\theta}{2}}$

So we have that :

$\displaystyle \lim_{\theta \to 0} \frac{\theta}{|u-v|}=\displaystyle \lim_{\theta \to 0} \frac{\theta}{2\cdot \sin {\frac{\theta}{2}}}=\displaystyle \lim_{\theta \to 0} \frac{\frac{\theta}{2}}{ \sin {\frac{\theta}{2}}}=\left(\displaystyle \lim_{\theta \to 0} \frac{ \sin {\frac{\theta}{2}}}{\frac{\theta}{2}}\right)^{-1}=1^{-1}=1$

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