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EAMCET Medical EAMCET Medical Solved Paper-1998

  • question_answer
    The vector \[P\to =a\widehat{i}+a\widehat{j}+3\widehat{k}\] and \[Q\to =a\widehat{i}+2\widehat{j}-\widehat{k}\]are perpendicular to each other, the positive value of a is:

    A)  3                                            

    B)  2

    C)  1                                            

    D)  zero

    Correct Answer: A

    Solution :

                     Since, vectors \[\vec{P}\]and \[\vec{Q}\] are perpendicular to each other, it means \[\vec{P}.\vec{Q}=P\,Q\,\cos \theta \]Here \[\theta ={{90}^{o}}\] So,          \[\vec{P}.\vec{Q}=0\] \[\therefore \]  \[(a\hat{i}+a\hat{j}+3\hat{k}).(a\hat{i}-2\hat{j}-\hat{k})=0\] \[\Rightarrow \]               \[{{a}^{2}}-2a-3=0\] \[{{a}^{2}}-3a+a-3=0\] \[a(a-3)+1(a-3)=0\] \[(a+1)(a-3)=0\] Therefore, positive value of \[a=3\]


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