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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    If three vectors a, b, c are such that \[a\ne 0\]and \[a\times b=2(a\times c),|a|=|c|=1,|b|=4\]and the angle between b and c is \[{{\cos }^{-1}}\left( \frac{1}{4} \right).\]Also \[b-2c=\lambda a,\]then find the value of \[\lambda \]

    A) \[~\pm \,4\]                                         

    B)         14                                         

    C)        \[\pm \text{ }2~\]                                

    D)         12

    Correct Answer: A

    Solution :

    \[a\times b=2(a\times c)\] \[\Rightarrow \]\[a\times (b-2c)=0\Rightarrow a\]a is parallel to \[b-2c.\]                 Now, \[(b-2c)=\lambda a\Rightarrow |b-2c{{|}^{2}}={{\lambda }^{2}}|a{{|}^{2}}\]                 \[\Rightarrow \]\[|b{{|}^{2}}+\,4|c{{|}^{2}}-4(b.c)={{\lambda }^{2}}|a{{|}^{2}}\]                 \[\Rightarrow \]\[16+4-4\times |b||c|\times \frac{1}{4}={{\lambda }^{2}}\]                 \[\Rightarrow \]\[20-4={{\lambda }^{2}}\Rightarrow \lambda =\pm 4\]


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