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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The resultant of two vectors\[\overrightarrow{P}\]and\[\overrightarrow{Q}\]is\[\overrightarrow{R}\]. If the magnitude of\[\overrightarrow{Q}\]is doubled, the new resultant becomes perpendicular to\[\overrightarrow{P}\]. Then the magnitude of\[\overrightarrow{R}\]is

    A)  \[P+Q\]                              

    B)  \[Q\]

    C)  \[P\]                                    

    D)  \[\frac{P+Q}{2}\]

    E)  \[P-Q\]

    Correct Answer: B

    Solution :

    In first case \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\]             ...(i) where\[\theta \]is the angle between P and Q. In second case,                 \[(\overrightarrow{P}+2\overrightarrow{Q}).\overrightarrow{P}=0\] or   \[\overrightarrow{P}\overrightarrow{P}+2(\overrightarrow{Q}.\overrightarrow{P})=0\] or         \[{{P}^{2}}+0=0\] or               \[P=0\] Putting\[P=0\]in Eq. (i), we get                 \[R=\sqrt{0+{{Q}^{2}}+0}=Q\]


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