Question

If $\theta$ is the angle between unit vectors $\stackrel{a}{\to }$ and $\stackrel{b}{\to }$, then $sin\left[\frac{\theta }{2}\right]$ equals

• A. $\frac{1}{2}|\stackrel{a}{\to }+\stackrel{b}{\to }|$
• B. $\frac{1}{2}|\stackrel{a}{\to }\times \stackrel{b}{\to }|$
• C. $\frac{1}{2}\sqrt{1-(\stackrel{a}{\to }.\stackrel{b}{\to })}$
• D. $\frac{1}{2}|\stackrel{a}{\to }-\stackrel{b}{\to }|$

Answer 1

The correct option is A $\frac{1}{2}|\stackrel{a}{\to }+\stackrel{b}{\to }|$

$\begin{array}{c}\text{Solution - According to the question}\overrightarrow{a}\text{and}\overrightarrow{b}\text{are}\\ \text{two unit vectors.}\\ \begin{array}{cc}\Rightarrow |\overrightarrow{a}|& =|\overrightarrow{b}|=1\\ |\overrightarrow{a}-\overrightarrow{b}{|}^2& =(a-b)(a+b)\\ & =|a{|}^2+|b{|}^2-2\left(|a||b|\cos \theta \right)\\ & =1+1-2\cos \theta \\ & =2-2\cos \theta \end{array}\end{array}$

We know that,

$1-\cos \theta =2{\sin }^2\frac{\theta }{2}$ $\Rightarrow |\overrightarrow{a}-\overrightarrow{b}{|}^2=2\times 2{\sin }^2\frac{\theta }{2}$ $\Rightarrow \sin \frac{\theta }{2}=\frac{1}{2}|\overrightarrow{a}-\overrightarrow{b}|$

Hence ,(A) is the correct option.