A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{2\pi }{3}\]
D) \[\pi \]
Correct Answer: C
Solution :
Given, a, b, c are unit vectors. \[\Rightarrow \]\[|a|=|b|=|c|=1\] ??. (i) Also, given \[a+b+c=0\]\[\Rightarrow \] \[(a+b)=-c\] Squaring on both sides, we get \[\Rightarrow \] \[{{(a+b)}^{2}}={{(c)}^{2}}\] \[\Rightarrow \] \[{{(a)}^{2}}+{{(b)}^{2}}+2a.b=|c{{|}^{2}}\] \[\Rightarrow \] \[|a{{|}^{2}}+|c{{|}^{2}}+2a.\,b=|c{{|}^{2}}\] \[[\because {{(a)}^{2}}=|a{{|}^{2}}]\] \[\Rightarrow \] \[1+1+2a.b=1\] [from Eq. (i) ] \[\Rightarrow \] \[2a\,.\,b=-1\] \[\Rightarrow \]\[a.b=-1/2=|a|\,|b|\,cos\theta \] \[\Rightarrow \] \[\cos \theta =-1/2=\cos 2\pi /3\] [from Eq. (i) ] \[\Rightarrow \] \[\theta =2\pi /3\]
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