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EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    Then the angle between \[\overrightarrow{A}\] and \[\overrightarrow{B}\] is: If \[\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C}\] and \[\left| \overrightarrow{A} \right.\left| = \right.\left| \overrightarrow{B} \right.=\left| \overrightarrow{C} \right.\left| {} \right.\]

    A) \[45{}^\circ \]                                   

    B) \[60{}^\circ \]

    C) \[90{}^\circ \]                                   

    D) \[120{}^\circ \]

    Correct Answer: D

    Solution :

                     \[{{C}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] Since,    \[A=B=C\] \[{{A}^{2}}={{A}^{2}}+{{A}^{2}}+2A\times Acos\theta \] \[{{A}^{2}}=2{{A}^{2}}(1+cos\theta )\] \[1+\cos \theta =\frac{1}{2}\] \[\cos \theta =\frac{1}{2}-1=-\frac{1}{2}=\cos {{120}^{o}}\] \[\theta ={{120}^{o}}\]


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