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JEE Main & Advanced Mathematics Vector Algebra Question Bank Modulus of vector Algebra of vectors

  • question_answer
    If a, b and c be three non-zero vectors, no two of which are collinear. If the vector \[\mathbf{a}+2\mathbf{b}\] is collinear with c and \[\mathbf{b}+3\mathbf{c}\] is collinear with a, then (\[\lambda \] being some non-zero scalar) \[\mathbf{a}+2\mathbf{b}+6\mathbf{c}\] is equal to                              [AIEEE 2004]

    A) \[\lambda \mathbf{a}\]        

    B) \[\lambda \mathbf{b}\]

    C) \[\lambda \mathbf{c}\]         

    D) 0

    Correct Answer: D

    Solution :

    Let \[\mathbf{a}+2\mathbf{b}=x\mathbf{c}\] and \[\mathbf{b}+3\mathbf{c}=y\mathbf{a},\] then \[\mathbf{a}+2\mathbf{b}+6\mathbf{c}=(x+6)\mathbf{c}\]and \[\mathbf{a}+2\mathbf{b}+6\mathbf{c}=(1+2y)\mathbf{a}\] So, \[(x+6)\mathbf{c}=(1+2y)\mathbf{a}\] Since \[\mathbf{a}\] and \[\mathbf{c}\] are non-zero and non-collinear, we have \[x+6=0\] and \[1+2y=0\] i.e., \[x=-6\] and \[y=-\frac{1}{2}.\] In either case, we have \[\mathbf{a}+2\mathbf{b}+6\mathbf{c}=\mathbf{0}\].


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