JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    Given two vectors are \[\hat{i}-\hat{j}\] and \[\hat{i}+2\hat{j}\], the unit vector coplanar with the two vectors and perpendicular to first is   AIEEE  Solved  Paper-2002

    A) \[\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})\]        

    B)           \[\frac{1}{\sqrt{5}}(2\hat{i}-\hat{j})\]       

    C)           \[\pm \frac{1}{\sqrt{2}}(\hat{i}-\hat{k})\]               

    D)           None of these

    Correct Answer: A

    Solution :

    Given, two vectors lie in \[xy\] -plane. Therefore, a vector coplanar with them is                                    \[a=x\,\hat{i}+y\,\hat{j}\] \[\therefore \]     \[a\bot (\hat{i}-\hat{j})\Rightarrow a.\,(\hat{i}-\hat{j})=0\] \[\Rightarrow \]   \[(x\hat{i}+y\hat{i}).\,(\hat{i}-\hat{j})=0\] \[\Rightarrow \]                   \[x-y=0\] \[\Rightarrow \]                   \[x=y\] \[\therefore \,a=x\,\hat{i}+x\,\hat{j}\] and \[\left| a \right|=\sqrt{{{x}^{2}}+{{x}^{2}}}=x\sqrt{2}\] \[\therefore \] Required unit vector \[=\frac{a}{\left| a \right|}=\frac{x(\hat{i}+\hat{j})}{x\sqrt{2}}\]                                    \[=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})\]


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