1
$\begingroup$

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

  • 0
    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!2012-01-13

2 Answers 2

1

Yes, your claim is correct....

  • 0
    Sorry, I wanted to delete my post but I couldn't.2012-01-13
7

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 :

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