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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1999

  • question_answer
    \[\vec{a}\] and \[\vec{b}\] are such that \[|\vec{a}+\vec{b}|\,=\,|\vec{a}-\vec{b}\] then angle between \[\vec{a}\] and \[\vec{b}\] is:

    A)  \[\pi /3\]            

    B)         \[\pi \]    

    C)  \[\pi /2\]            

    D)         zero

    Correct Answer: C

    Solution :

    \[|\vec{a}+\vec{b}=\,|\vec{a}-\vec{b}|\]                 \[\therefore \]  \[{{a}^{2}}+{{b}^{2}}+2|\vec{a}|\vec{b}|\,\cos \,\theta \] \[={{a}^{2}}+{{b}^{2}}-2\,|\vec{a}|\,|\vec{b}|\,\cos \theta \] On solving \[\theta =\frac{\pi }{2}\].


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