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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    The resultant of two vectors \[\vec{A}\] and \[\vec{B}\] is perpendicular to the vector \[\vec{A}\] and its magnitude is equal to half the magnitude of vector \[\vec{B}\]. The angle between \[\vec{A}\] and \[\vec{B}\] is

    A) \[120{}^\circ \]  

    B)        \[150{}^\circ \]

    C) \[135{}^\circ \]

    D)        None of these

    Correct Answer: B

    Solution :

     \[\frac{B}{2}=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]     ......(i) \[\therefore \,\,\tan 90{}^\circ =\frac{B\sin \theta }{A+B\cos \theta }\Rightarrow A+B\cos \theta =0\] \[\therefore \,\,\cos \theta =-\frac{A}{B}\] Hence, from (i) \[\frac{{{B}^{2}}}{4}={{A}^{2}}+{{B}^{2}}-2{{A}^{2}}\Rightarrow A=\sqrt{3}\frac{B}{2}\] \[\Rightarrow \,\,\cos \theta =\frac{A}{B}=-\frac{\sqrt{3}}{2}\] \[\therefore \,\,\theta =150{}^\circ \]


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