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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    The vectors \[\text{\vec{A}}\] and \[\text{\vec{B}}\] are such that\[\text{ }\!\!|\!\!\text{ }\overrightarrow{\text{A}}\text{+}\overrightarrow{\text{B}}\text{ }\!\!|\!\!\text{ }\,\text{=}\,\text{ }\!\!|\!\!\text{ }\overrightarrow{\text{A}}\text{-}\overrightarrow{\text{B}}|\] The angle between the two vectors is

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

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

    C) \[75{}^\circ \]

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

    Correct Answer: A

    Solution :

                    As we have given \[|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|\] or            \[\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]                                 \[=\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\,\cos \theta }\] where \[\theta \] is the angle between \[\overrightarrow{A}\] and \[\overrightarrow{B}\]. Squaring both sides, we have \[{{A}^{2}}+{{B}^{2}}+2\,AB\,\cos \theta ={{A}^{2}}+{{B}^{2}}-2AB\,\cos \theta \] or            \[4AB\,\,\cos \theta =0\] As           \[AB\ne 0\] \[\therefore \]  \[\cos \theta =0-\cos {{90}^{o}}\]             \[\therefore \]  \[\theta ={{90}^{o}}\] Hence, angle between \[\overrightarrow{A}\] and \[\overrightarrow{B}\] is 90°.


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