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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 41

  • question_answer
    If \[\vec{A}\] and \[\vec{B}\] are two non-zero vectors such that \[\left| \vec{A}+\vec{B} \right|=\frac{\left| \vec{A}+\vec{B} \right|}{2}\] and \[\left| {\vec{A}} \right|=2\left| {\vec{B}} \right|,\] then the angle between \[\vec{A}\] and \[\vec{B}\] is

    A) \[37{}^\circ \]                  

    B)        \[53{}^\circ \]                  

    C) \[{{\cos }^{-1}}\left( -3/4 \right)\]     

    D)        \[{{\cos }^{-1}}\left( -5/12 \right)\]

    Correct Answer: C

    Solution :

    [c]  \[\left| \vec{A}+\vec{B} \right|=\frac{\left| \vec{A}-\vec{B} \right|}{2}\] \[{{\left| \vec{A}+\vec{B} \right|}^{2}}=\frac{{{\left| \vec{A}-\vec{B} \right|}^{2}}}{4}\] \[\left( {{A}^{2}}+{{B}^{2}}+2AB\cos \theta  \right)=\frac{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta }{4}\] \[\left| {\vec{A}} \right|\,=\,\left| {\vec{B}} \right|\Rightarrow {{A}^{2}}=4{{B}^{2}}\] \[4[4{{B}^{2}}+{{B}^{2}}+2(2B)(B)\cos \theta ]\] \[=4{{B}^{2}}+{{B}^{2}}-2(2B)(B)\cos \theta \] \[4\left( 5{{B}^{2}}+4{{B}^{2}}\cos \theta  \right)=5{{B}^{2}}-4{{B}^{2}}\cos \theta \] \[20\cos \theta =-15\] \[\cos \theta =\frac{-3}{4}\]


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