Browse Questions

# If $\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$ ,than prove that tan$\large\frac{\theta}{2}$=$\large\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$

Toolbox:
• For any two vectors $\hat a \: and \: \hat b$ $(\hat a + \hat b)^2=(\hat a)^2+2\hat a.\hat b+(\hat b)^2=a^2+2\hat a.\hat b+b^2$ $(\hat a-\hat b)^2=a^2-2\hat a.\hat b+b^2$ $(\hat a+\hat b).(\hat a-\hat b)=a^2-b^2$
Step 1
$|\hat a|=1,\: |\hat b|=1$
$\large\frac{|\hat a-\hat b|^2}{\hat a+\hat b|^2} = \large\frac{(\hat a-\hat b)^2}{(\hat a+\hat b)^2} = \large\frac{\hat a^2-2\hat a.\hat b+\hat b^2}{\hat a^2+2\hat a.\hat b+\hat b^2}$
$= \large\frac{1-2(1)(1) \cos\theta+1}{1+2(1)(1) \cos\theta+1}$
$= \large\frac{2(1-\cos\theta)}{2(1+\cos\theta)}$
$= \large\frac{2\sin^2\large\frac{\theta}{2}}{2\cos^2\large\frac{\theta}{2}}$
$= \tan^2 \large\frac{\theta}{2}$
Step 2
$\therefore \large\frac{|\hat a-\hat b|}{| \hat a + \hat b|}=tan \large\frac{\theta}{2}$