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

  • question_answer
    If \[4\mathbf{i}+7\mathbf{j}+8\mathbf{k},\,\,\,2\mathbf{i}+3\mathbf{j}+4\mathbf{k}\,\] and \[2\mathbf{i}+5\mathbf{j}+7\mathbf{k}\] are the position vectors of the vertices A, B and C respectively of triangle ABC. The position vector of the point where the bisector of angle A meets BC is        [Pb. CET 2004]

    A) \[\frac{1}{3}\,(6\mathbf{i}+13\mathbf{j}+18\mathbf{k})\]         

    B) \[\frac{2}{3}\,(6\mathbf{i}+12\mathbf{j}-8\mathbf{k})\]

    C) \[\frac{1}{3}\,(-6\mathbf{i}-8\mathbf{j}-9\mathbf{k})\]

    D) \[\frac{2}{3}\,(-6\mathbf{i}-12\mathbf{j}+8\mathbf{k})\]

    Correct Answer: A

    Solution :

    Let the bisector of angle A meets BC at D, then AD divides BC in the ratio AB: AC  Position vectors of D =\[\frac{|\overrightarrow{AB}|(2\mathbf{i}+5\mathbf{j}+7\mathbf{k})+|\overrightarrow{AC}|(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})}{|\overrightarrow{AB}|+|\overrightarrow{AC}|}\] Here, \[|\overrightarrow{AB}|=|-2\mathbf{i}-4\mathbf{j}-4\mathbf{k}|=6\] and \[|\overrightarrow{AC}|=|-2\mathbf{i}-2\mathbf{j}-\mathbf{k}|=3\] Position vector of \[D=\frac{6(2\mathbf{i}+5\mathbf{j}+7\mathbf{k})+3(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})}{6+3}\]                   \[=\frac{18\mathbf{i}+39\mathbf{j}+54\mathbf{k})}{9}\]\[=\frac{1}{3}(6\mathbf{i}+13\mathbf{j}+18\mathbf{k})\].


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