A) \[\pi /6\]
B) \[\pi /3\]
C) \[\pi /2\]
D) \[2\pi /3\]
Correct Answer: D
Solution :
Given condition is \[\mathbf{a}+\mathbf{b}=\mathbf{c}.\] Using dot product, \[(\mathbf{a}+\mathbf{b}).(\mathbf{a}+\mathbf{b})=\mathbf{c}.\mathbf{c}\] \[\Rightarrow \mathbf{a}.\mathbf{a}+\mathbf{b}.\mathbf{b}+2\mathbf{a}.\mathbf{b}=\mathbf{c}.\mathbf{c}\] \[\Rightarrow \,|\mathbf{a}|.|\mathbf{a}|\cos 0{}^\circ +|\mathbf{b}|.|\mathbf{b}|\cos 0{}^\circ +2|\mathbf{a}|.|\mathbf{b}|\cos \alpha \] \[=\,|\mathbf{c}|.|\mathbf{c}|\cos 0{}^\circ \], \[(\because \,\,\,|\mathbf{a}|\,=\,|\mathbf{b}|\,=\,|\mathbf{c}|\,=1)\] \[\Rightarrow 1+1+2\cos \alpha =1\Rightarrow \cos \alpha =-\frac{1}{2}\Rightarrow \alpha =\frac{2\pi }{3}\].
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