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  • question_answer
    If \[\vec{A}\] and \[\vec{B}\] are two vectors such that\[|\vec{A}+\vec{B}|\,=\,|\vec{A}-\vec{B}|,\] the angle between vectors \[\vec{A}\] and \[\vec{B}\] is: [AIPMT 2001]

    A) \[{{0}^{o}}\]

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

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

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

    Correct Answer: C

    Solution :

    We have given
                            \[|\vec{A}+\vec{B}|=\,|\vec{A}-\vec{B}|\]
                Squaring both the sides, we obtain
                            \[{{\left| \vec{A}+\vec{B} \right|}^{2}}=\,{{\left| \vec{A}-\vec{B} \right|}^{2}}\]
    or         \[(\vec{A}+\vec{B})\,.\,(\vec{A}+\vec{B})\,=(\vec{A}-\vec{B})\,.(\vec{A}-\vec{B})\]
    or         \[\vec{A}.\vec{A}+\vec{A}.\vec{B}+\vec{B}.\vec{A}+\vec{B}.\vec{B}=\vec{A}.\vec{A}-\vec{A}.\vec{B}\]\[-\vec{B}\,.\,\vec{A}+\vec{B}\,.\,\vec{B}\]
    or         \[\vec{A}.\vec{B}+\vec{A}.\vec{B}=-\vec{A}.\vec{B}-\vec{A}.\vec{B}\]\[(\because \,\,\vec{B}.\vec{A}=\vec{A}.\vec{B})\]
    or         \[4\vec{A}.\vec{B}=0\]
    or         \[\vec{A}\,.\,\vec{B}=0\]
    Since dot product of \[\vec{A}\] and \[\vec{B}\] is zero hence, \[\vec{A}\]and \[\vec{B}\] are mutually perpendicular i.e., angle between and \[\vec{A}\] is \[\vec{B}\]\[90{}^\circ \]


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