Browse Questions

If $\hat{a}\;and\;\hat{b}$ are unit vectors inclined at an angle $\theta$,then prove that$(a)\;\cos \frac{\theta}{2}=\frac{1}{2}|\hat{a}+\hat{b}|\qquad(b)\;\tan \frac{\theta}{2}=\frac {|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$

Toolbox:
• $\large|\overrightarrow a+\overrightarrow b|^2=(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)$
• $\large(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b$
• $\large\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$
• $\large\:1+cos\theta=2cos^2\frac{\theta}{2}$
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 $\large\:1+cos\theta=2cos^2\frac{\theta}{2}$
$\Rightarrow \large|\overrightarrow a+\overrightarrow b|^2=2.2cos^2\frac{\theta}{2}$
Taking squareroot on both the sides we get the result
$\large|\overrightarrow a+\overrightarrow b|=2cos\frac{\theta}{2}$
$\Rightarrow \large\:\frac{1}{2}cos\frac{\theta}{2}=|\overrightarrow a+\overrightarrow b|$
Hence proved.