Question

If $ \vec{x}+\vec{y}+\vec{z}=\vec{0},\,|\vec{x}|=|\vec{y}|=|\vec{z}|=2 $ and $ \theta $ is angle between $ \vec{y} $ and $ \vec{z}, $ then the value of $ \text{cose}{{\text{c}}^{2}}\theta +{{\cot }^{2}}\theta $ is equal to

• A. $ 4/3 $
• B. $ 5/3 $
• C. $ 1/3 $
• D. $ 1 $

Answer 1

Answer: (B)

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} $