is:
A) \[2sin\theta \]
B) \[2.sin\frac{\theta }{2}\]
C) \[cos\theta \]
D) \[\frac{1}{2}.cos\frac{\theta }{2}\]
Correct Answer: B
Solution :
[b] Since, we know that \[{{(\vec{a}-\vec{b})}^{2}}={{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}-2\vec{a}.\vec{b}\] \[=1+1-2ab.cos\theta =2\left( 1-cos\theta \right)=2.2si{{n}^{2}}\frac{\theta }{2}\] \[\Rightarrow \left| {{(\vec{a}-\vec{b})}^{2}} \right|=4si{{n}^{2}}\frac{\theta }{2}\] \[\Rightarrow \left| \vec{a}-\vec{b} \right|=2.sin\frac{\theta }{2}\] Hence, option [b] is correct.
You need to login to perform this action.
You will be redirected in
3 sec
