A) \[{{90}^{o}}\]
B) \[{{180}^{o}}\]
C) \[{{120}^{o}}\]
D) zero
Correct Answer: D
Solution :
We have, \[\vec{a}+\vec{b}=\vec{c}\]and \[c=a+b\] \[\Rightarrow \] \[c=\sqrt{{{a}^{2}}+{{b}^{2}}+2ab\,\cos \theta }\] \[\Rightarrow \] \[a+b=\sqrt{{{a}^{2}}+{{b}^{2}}+2ab\,\cos \theta }\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}+2ab={{a}^{2}}+{{b}^{2}}+2ab\,\cos \theta \] \[\Rightarrow \] \[\cos \theta =1\] \[\Rightarrow \] \[\theta ={{0}^{o}}\]
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