# If $\large\overrightarrow a\:and\:\overrightarrow b$ are unit vectors then prove that $\large\frac{1}{2}|\overrightarrow a-\overrightarrow b|=sin\frac{\theta}{2}.$

Toolbox:
• $\large|\overrightarrow a-\overrightarrow b|^2=(\overrightarrow a-\overrightarrow b).(\overrightarrow a-\overrightarrow b)$
• $\large(\overrightarrow a-\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2-2\overrightarrow a.\overrightarrow b$
• $\large\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$
• $\large\:1-cos\theta=2sin^2\frac{\theta}{2}$
Given that $\large\overrightarrow a\:\:and\:\:\overrightarrow b$ are unit vectors.
$\Rightarrow\:\large|\overrightarrow a|=|\overrightarrow b|=1$
We know that $\large|\overrightarrow a-\overrightarrow b|^2=(\overrightarrow a-\overrightarrow b).(\overrightarrow a-\overrightarrow b)$
$=\large|\overrightarrow a|^2+|\overrightarrow b|^2-2\overrightarrow a.\overrightarrow b$
$=\large|\overrightarrow a|^2+|\overrightarrow b|^2-2|\overrightarrow a||\overrightarrow b|cos\theta$
But it is given that $\large|\overrightarrow a|=|\overrightarrow b|=1$
$\Rightarrow\large|\overrightarrow a+\overrightarrow b|^2=1+1-2cos\theta=2-2cos\theta$
$\large=2(1-cos\theta)$
We know that $\large\:1-cos\theta=2sin^2\frac{\theta}{2}$
$\Rightarrow \large|\overrightarrow a-\overrightarrow b|^2=2.2sin^2\frac{\theta}{2}$
Taking squareroot on both the sides we get the result
$\large|\overrightarrow a-\overrightarrow b|=2sin\frac{\theta}{2}$
$\Rightarrow \large\:\frac{1}{2}sin\frac{\theta}{2}=|\overrightarrow a-\overrightarrow b|$
Hence proved.

edited Apr 19, 2013