Question

If $e_1$ and $e_2$ the two unit vectors $\theta$ is the angle between them, then prove that $sin\left(\frac{\theta }{2}\right)$ is $\frac{1}{2}|e1-e2|$

Answer 1

$|A-B|=\sqrt{(|A|²+|B|²-2|A||B|.cos\theta )}$

$|e1-e2|=\sqrt{(|e1|²+|e2|²-2|e1|.|e2|cos\theta )}$

here,$e1$ and $e2$ unit vectors

so,$|e1|=1,|e2|=1$

$|e1-e2|=\sqrt{(1²+1²-\mathrm{2.1.1}cos\theta )}$

$=\sqrt{(2-2cos\theta )}$

$=\sqrt{2(1-cos\theta )}$

$=\sqrt{\frac{2\ast 2sin²\theta }{2}}$

$=\frac{2sin\theta }{2}$

hence,$|e1-e2|=\frac{2sin\theta }{2}$

$sin\frac{\theta }{2}=\frac{1}{2}|e1-e2|$

hence proved.