A) \[\frac{1}{2}|\mathbf{a}-\mathbf{b}|\]
B) \[\frac{1}{2}|\mathbf{a}+\mathbf{b}|\]
C) \[\frac{|\mathbf{a}-\mathbf{b}|}{|\mathbf{a}+\mathbf{b}|}\]
D) \[\frac{|\mathbf{a}+\mathbf{b}|}{|\mathbf{a}-\mathbf{b}|}\]
Correct Answer: B
Solution :
\[(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}+\mathbf{b})=|\mathbf{a}{{|}^{2}}+|{{\mathbf{b}}^{2}}|+2\mathbf{a}\cdot \mathbf{b}\] \[=1+1+2(1)(1)\cos \theta =2+2\cos \theta \] \[\Rightarrow \] \[|\mathbf{a}+\mathbf{b}{{|}^{2}}=2\cdot 2\cos \frac{\theta }{2}\] \[\Rightarrow \] \[\cos \frac{\theta }{2}=\frac{1}{2}|\mathbf{a}+\mathbf{b}|\]
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