JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    The vector \[\widehat{\mathbf{i}}+x\widehat{\mathbf{j}}+3\widehat{\mathbf{k}}\] is rotated through an angle \[0\] and doubled in magnitude, then it becomes\[4\widehat{\mathbf{i}}+(4x-2)\widehat{\mathbf{j}}+2\widehat{\mathbf{k}}\]. The value of \[x\] is

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

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

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

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

    Correct Answer: A

    Solution :

    We have,\[2|\widehat{\mathbf{i}}+x\widehat{\mathbf{j}}+3\widehat{\mathbf{k}}|\]                                 \[=|4\widehat{\mathbf{i}}+(4x-2)\widehat{\mathbf{j}}+2\widehat{\mathbf{k}}|\] \[\Rightarrow \]               \[2\sqrt{1+{{x}^{2}}+9}=\sqrt{{{4}^{2}}{{(4x-2)}^{2}}+{{2}^{2}}}\] \[\Rightarrow \]               \[4({{x}^{2}}+10)=16{{x}^{2}}+4-16x+4\] \[\Rightarrow \]               \[12{{x}^{2}}-16x-16=0\] \[\Rightarrow \]               \[3{{x}^{2}}-4x-4=0\] \[\Rightarrow \]               \[(3x+2)(x-2)=0\] \[\Rightarrow \]               \[x=2,\,\,-\frac{2}{3}\]


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