If points $D, E, F$ divide the sides $BC, CA, AB$ of $\triangle ABC$ in the ratios $1:4, 3:2, 3:7$ respectively,and point $K$ divides $AB$ in some ratio,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK} = ......$

  • A
    $1:1$
  • B
    $2:5$
  • C
    $5:2$
  • D
    None of these

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$A$. Unit vector in the direction opposite to that $a-b$ is$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
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Let $\vec{a}_n = (\tan \theta_n)\hat{i} + \hat{j}$ and $\vec{b}_n = \hat{i} - (\cot \theta_n)\hat{j}$, where $\theta_n = \frac{2^{n-1}\pi}{2^n+1}$, for some $n \in N, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a}_k|^2}{\sum_{k=1}^n |\vec{b}_k|^2}$ is . . . . . . .

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