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 :
\[(\mathbf{a}+\mathbf{b}).(\mathbf{a}+\mathbf{b})=\,|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+\,2\mathbf{a}\,.\,\mathbf{b}\] or \[|\mathbf{a}+\mathbf{b}{{|}^{2}}=2.2{{\cos }^{2}}\frac{\theta }{2}\Rightarrow \cos \frac{\theta }{2}=\frac{1}{2}|\mathbf{a}+\mathbf{b}|.\]
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