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

  • question_answer
    If ABCD is a parallelogram, \[\overrightarrow{AB}=2\,\mathbf{i}+4\,\mathbf{j}-5\,\mathbf{k}\] and \[\overrightarrow{AD}=\,\mathbf{i}+2\,\mathbf{j}+3\,\mathbf{k},\] then the unit vector in the direction of BD is                                                                                                               [Roorkee 1976]

    A) \[\frac{1}{\sqrt{69}}\,(\mathbf{i}+2\mathbf{j}-8\mathbf{k})\]    

    B) \[\frac{1}{69}\,(\mathbf{i}+2\mathbf{j}-8\,\mathbf{k})\]

    C) \[\frac{1}{\sqrt{69}}\,(-\mathbf{i}-2\mathbf{j}+8\mathbf{k})\]

    D) \[\frac{1}{69}\,(-\mathbf{i}-2\mathbf{j}+8\,\mathbf{k})\]

    Correct Answer: C

    Solution :

    Since \[\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD}\Rightarrow \overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB}\]                 \[=(\mathbf{i}+2\mathbf{j}+3\mathbf{k})-(2\mathbf{i}+4\mathbf{j}-5\mathbf{k})=-\mathbf{i}-2\mathbf{j}+8\mathbf{k}\] Hence unit vector in the direction of \[\overrightarrow{BD}\] is \[\frac{-\mathbf{i}-2\mathbf{j}+8\mathbf{k}}{|-\mathbf{i}-2\mathbf{j}+8\mathbf{k}|}=\frac{-\mathbf{i}-2\mathbf{j}+8\mathbf{k}}{\sqrt{69}}.\]         


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