Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers 2

up vote 1 down vote accepted

Yes, your claim is correct....

share|improve this answer
    
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
3  
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
add comment

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 :

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.