A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) 0
D) None of these
Correct Answer: C
Solution :
\[\mathbf{a}+\mathbf{b}=\mathbf{c}\Rightarrow \,|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+2\mathbf{a}\,.\,\mathbf{b}=\,|\mathbf{c}{{|}^{2}}\] and \[|\mathbf{a}|+|\mathbf{b}|\,=\,|\mathbf{c}|\] \[\Rightarrow \,|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+2|\mathbf{a}||\mathbf{b}|\,=\,|\mathbf{c}{{|}^{2}}\] \[\therefore \,\mathbf{a}\,.\,\mathbf{b}=\,|\mathbf{a}||\mathbf{b}|\Rightarrow \cos \theta =1\] Þ \[\theta =0.\]
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