CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If \[|\overrightarrow{a}|=|\overrightarrow{b}|=1\]and\[|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{3},\]then the value of \[(3\overrightarrow{a}-4\overrightarrow{b}).(2\overrightarrow{a}+5\overrightarrow{b})\]is:

    A)  \[-21\]                                

    B)  \[-\frac{21}{2}\]

    C)  21                         

    D)         \[\frac{21}{2}\]

    E)  \[\frac{59}{2}\]          

    Correct Answer: B

    Solution :

    \[\because \] \[|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{3}\] \[\Rightarrow \]                 \[|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}+2\overrightarrow{a}.\overrightarrow{b}=3\] \[\Rightarrow \]               \[\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}\] \[\therefore \]  \[(3\overrightarrow{a}-4\overrightarrow{b}).(2\overrightarrow{a}+5\overrightarrow{b})\] \[=6|\overrightarrow{a}{{|}^{2}}+15\overrightarrow{a}.\overrightarrow{b}-8\overrightarrow{a}.\overrightarrow{b}-20|\overrightarrow{b}{{|}^{2}}\] \[=6+7\overrightarrow{a}.\overrightarrow{b}-20\] \[=6+\frac{7}{2}-20=\frac{7-28}{2}=-\frac{21}{2}\]


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