JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    The vector \[\hat{i}+x\hat{j}+3\hat{k}\] is rotated through an angle \[\theta \] and doubled in magnitude, then it becomes \[4\hat{i}+(4x-2)\hat{j}+2\hat{k}\]. The values of \[x\] are   AIEEE  Solved  Paper-2002

    A) \[\left\{ -\frac{2}{3},2 \right\}\]

    B)           \[\left\{ \frac{1}{3},2 \right\}\]     

    C)           \[\left\{ \frac{2}{3},0 \right\}\]     

    D)           \[\left\{ 2,7 \right\}\]

    Correct Answer: A

    Solution :

       Since, the vector \[\hat{i}+x\hat{j}+3\hat{k}\] doubled in magnitude, then it becomes\[4\hat{i}+(4x-2)\hat{j}+2\hat{k}\]. \[\therefore \]     \[3\left| \hat{i}+x\,\hat{j}+3\hat{k} \right|=\left| 4\,\hat{i}+(4x-2)\,\hat{j}+2\hat{k} \right|\] \[\Rightarrow \]   \[2\sqrt{1+{{x}^{2}}+9}=\sqrt{16+{{(4x-2)}^{2}}+4}\] \[\Rightarrow \]   \[40+4{{x}^{2}}=20+{{(4x-2)}^{2}}\] \[\Rightarrow \,12{{x}^{2}}-16x-16=0\] \[\Rightarrow \]   \[3{{x}^{2}}-4x-4=0\] \[\Rightarrow \]   \[(x-2)\,(3x+2)=0\] \[\Rightarrow \]   \[x=2,-\frac{2}{3}\]


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