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JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    The points D, E, F divide BC, CA and AB of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio \[1:3\], then \[(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF})\,\,:\,\,\overrightarrow{CK}\] is equal to [MNR 1987]

    A) 1 : 1                                          

    B) 2 : 5

    C) 5 : 2                                          

    D) None of these

    Correct Answer: B

    Solution :

    • Let \[\overrightarrow{AB}=\mathbf{a},\] \[\overrightarrow{AC}=\mathbf{b}\]                     
    • So,\[\overrightarrow{AD}=\frac{4\mathbf{a}+\mathbf{b}}{5},\]\[\overrightarrow{AE}=\frac{2\mathbf{b}}{5},\] \[\overrightarrow{AF}=\frac{3\mathbf{a}}{10},\] and \[\overrightarrow{AK}=\frac{\mathbf{a}}{4}\]
    • \[\frac{\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}}{\overrightarrow{CK}}=\frac{\frac{\mathbf{b}+4\mathbf{a}}{5}+\frac{2\mathbf{b}-5\mathbf{a}}{5}+\frac{3\mathbf{a}-10\mathbf{b}}{10}}{\frac{\mathbf{a}-4\mathbf{b}}{4}}\]                                                                          
    • \[=\frac{6\mathbf{b}-2\mathbf{a}+3\mathbf{a}-10\mathbf{b}}{10(\mathbf{a}-4\mathbf{b})}\times 4=\frac{2}{5}\].


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