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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    \[\mathbf{a}\]and \[\mathbf{c}\] are unit vectors and\[|\mathbf{b}|\,\,=4\]. If angle between \[\mathbf{b}\] and \[\mathbf{c}\] is and \[{{\cos }^{-1}}\left( \frac{1}{4} \right)\]\[\text{and}\]\[\mathbf{a}\times \mathbf{b}=2\mathbf{a}\times \mathbf{c}\], then\[\mathbf{b}=\lambda \mathbf{a}+2\mathbf{c}\], where\[\lambda \]is equal to

    A) \[\pm \frac{1}{4}\]

    B) \[\pm \frac{1}{2}\]

    C) \[\pm \,\,1\]

    D)  None of the above

    Correct Answer: C

    Solution :

    \[\mathbf{a}\times \mathbf{b}=2\mathbf{a}\times \mathbf{c}\] \[\Rightarrow \]               \[\mathbf{a}\times \mathbf{b}-2\mathbf{a}\times \mathbf{c}=0\] \[\Rightarrow \]               \[\mathbf{a}\times (\mathbf{b}-2\mathbf{c})=0\] \[\Rightarrow \]               \[\mathbf{a}\]and\[(\mathbf{b}-2\mathbf{c})\]are collinear. \[\Rightarrow \]               \[\mathbf{b}-2\mathbf{c}=\lambda \mathbf{a}\] \[\Rightarrow \]               \[|\mathbf{b}-2\mathbf{c}{{|}^{2}}={{\lambda }^{2}}|\mathbf{a}{{|}^{2}}\] \[\Rightarrow \]               \[|\mathbf{b}{{|}^{2}}+4|\mathbf{c}{{|}^{2}}-4\,\,\mathbf{b}\cdot \mathbf{c}={{\lambda }^{2}}|\mathbf{a}{{|}^{2}}\] \[\Rightarrow \]               \[16+4-4(4)(1)\frac{1}{4}={{\lambda }^{2}}(1)\] \[\therefore \]  \[\lambda =\pm 4\]


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