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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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The vector $\hat i+x\hat j+3\hat k$ is rotated about angle $\theta$ and is doubled in magnitude, then it becomes $4\hat i+(4x-2)\hat j+2\hat k$. The value of $x=?$

$\begin{array}{1 1} (\frac{-2}{3},2) \\(\frac{1}{3},2) \\ (\frac{2}{3},0) \\ (2,7)\end{array} $

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Given magnitude of $4\hat i+(4x-2)\hat j+2\hat k$ is double that of $\hat i+x\hat j+3\hat k$
$\therefore \:40+4x^2=20+(4x-2)^2$
or $\:3x^2-4x-4=0$
answered Nov 4, 2013 by rvidyagovindarajan_1

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