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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    Let \[\vec{a},\vec{b},\vec{c}\]be three non-zero vectors which are pairwise non-collinear. If \[\vec{a}+3\vec{b}\]is collinear with \[\vec{c}\] and \[\vec{b}+2\vec{c}\]is collinear with \[\vec{a}+3\vec{b}+6\vec{c}\] is:     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[\vec{a}\]

    B)  \[\vec{c}\]

    C) \[\vec{0}\]

    D)  \[\vec{a}+\vec{c}\]

    Correct Answer: C

    Solution :

                    \[\vec{a}+3\vec{b}=\lambda \vec{c}\]                                                                                                                   ??(1) \[\vec{b}+2\vec{c}=\mu \vec{a}\]                                                                                                                           ??(2) (1)-3(2) gives \[(1+3\mu )\vec{a}-(\lambda +6)\vec{c}=0\] As \[\vec{a}\] and \[\vec{c}\] are non collinear \[\therefore \]\[1+3\mu =0\]and\[\lambda +6=0\] From\[(1)\]\[\vec{a}+3\vec{b}+6\vec{c}=\vec{0}\]


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