A) \[4/3\]
B) \[5/3\]
C) \[1/3\]
D) \[1\]
Correct Answer: B
Solution :
Given, \[\vec{x}+\vec{y}+\vec{z}=\vec{0}\] and \[|\vec{x}|=|\vec{y}|=|\vec{z}|=2\] \[\therefore \] \[\vec{x}=-\vec{y}-\vec{z}\] \[\Rightarrow \] \[{{\vec{x}}^{2}}={{(\vec{y}+\vec{z})}^{2}}\] \[\Rightarrow \] \[|\vec{x}{{|}^{2}}=|\vec{y}{{|}^{2}}+|\vec{z}{{|}^{2}}+2|y||z|\,\cos \,\theta \] \[\Rightarrow \] \[4=4+4+2\times 2\times 2\cos \theta \] \[\Rightarrow \] \[\cos \theta =-\frac{4}{8}=-\frac{1}{2}=\cos \,{{120}^{o}}\] \[\Rightarrow \] \[\theta ={{120}^{o}}\] Now, \[\text{cose}{{\text{c}}^{2}}\theta +{{\cot }^{2}}\theta =\text{cose}{{\text{c}}^{2}}{{120}^{o}}+{{\cot }^{2}}{{120}^{o}}\] \[={{\left( \frac{2}{\sqrt{3}} \right)}^{2}}+{{\left( -\frac{1}{\sqrt{3}} \right)}^{2}}\] \[=\frac{4}{3}+\frac{1}{3}=\frac{5}{3}\]
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