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JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • question_answer
    If three vectors \[\vec{a},\vec{b},\vec{c}\] are such that \[\vec{a},\ne 0\] and \[\vec{a}\times \vec{b}=2\vec{a}\times \vec{c}\], \[\left| {\vec{a}} \right|=\left| {\vec{b}} \right|=1,\,\left| {\vec{b}} \right|=4\] and the angle between \[\vec{b},\,\vec{c}\] is \[{{\cos }^{-1}}\frac{1}{4}\] then \[\vec{b}-2\vec{c}=\lambda \vec{a}\] where \[\lambda (\lambda >0)\] is equal to

    A)  2                                

    B)  4

    C)  \[\frac{1}{2}\]                         

    D)  \[\frac{1}{4}\]

    Correct Answer: B

    Solution :

    \[\vec{a}\times \vec{b}\,=2\vec{a}\times \vec{c}\Rightarrow \vec{a}\times (\vec{b}-2\vec{c})=0\] \[\Rightarrow \,\vec{b}-2\vec{c}=\lambda \vec{a}\] But \[{{(\vec{b}-2\vec{c})}^{2}}\,={{\vec{b}}^{2}}+4{{\vec{c}}^{2}}-4\vec{b}.\vec{c}\] \[=16\,+4-4\times 4\,\times \,\frac{1}{4}=16\] Hence, \[{{\lambda }^{2}}{{a}^{2}}=16\Rightarrow \lambda =4\].


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