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BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    If the angle between the unit vectors \[\vec{a}\]and  \[\vec{b}\]be \[{{60}^{o}},\] then \[(2\vec{a}-3\vec{b}).(4\vec{a}+\vec{b})\]is equal to:

    A)  0                            

    B)  5                            

    C)  15                         

    D)         none of these

    Correct Answer: A

    Solution :

    Since, the angle between the vectors is \[{{60}^{o}}\] \[\therefore \]  \[\vec{a}.\vec{b}=|\vec{a}||\vec{b}|cos{{60}^{o}}\]                 \[=1.1.\frac{1}{2}\]          \[\left( \therefore \,|\vec{a}|=|\vec{b}|=1 \right)\]                 \[=\frac{1}{2}\]                                 ?(i) Now,     \[(2\vec{a}-3\vec{b}).(4\vec{a}+\vec{b})\] \[=8\vec{a}.\vec{a}+2\vec{a}.\vec{b}-12\vec{a}.\vec{b}-3\vec{b}.\vec{b}\]                 \[=8(1)-10\vec{a}.\vec{b}-3(1)\]                               \[[\because \,\,\vec{a}.\vec{a}=1]\]                                 \[=5-10\left( \frac{1}{2} \right)\]                               [from (i)]                                 \[=0\]


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