A) \[\frac{1}{2}|{{\vec{e}}_{1}}+{{\vec{e}}_{2}}|\]
B) \[\frac{1}{2}|{{\vec{e}}_{1}}+{{\vec{e}}_{2}}|\]
C) \[\frac{{{{\vec{e}}}_{1}}.{{{\vec{e}}}_{2}}}{2}\]
D) \[\frac{|{{{\vec{e}}}_{1}}\times {{{\vec{e}}}_{2}}|}{2|{{{\vec{e}}}_{1}}||{{{\vec{e}}}_{2}}|}\]
Correct Answer: B
Solution :
\[\text{Consider }\!\!|\!\!\text{ }{{\hat{e}}_{1}}-{{\hat{e}}_{2}}{{|}^{2}}=2-2\cos \theta =4{{\sin }^{2}}\frac{\theta }{2}\] \[\therefore \] \[\frac{1}{2}|{{\hat{e}}_{1}}-{{\hat{e}}_{2}}|=\sin \frac{\theta }{2}\]
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