The points D, E, F divide BC, CA and AB of th | vedclass

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(B) Let $\overrightarrow{AB} = \vec{c}$ and $\overrightarrow{AC} = \vec{b}$.Then $\overrightarrow{AD} = \frac{4\overrightarrow{AB} + 1\overrightarrow{

Vedclass · Accepted Answer

$2:5$

Vedclass · Answer

(B) Let $\overrightarrow{AB} = \vec{c}$ and $\overrightarrow{AC} = \vec{b}$.Then $\overrightarrow{AD} = \frac{4\overrightarrow{AB} + 1\overrightarrow{AC}}{1+4} = \frac{4\vec{c} + \vec{b}}{5}$.Since $E$ divides $CA$ in ratio $3:2$,$\overrightarrow{AE} = \frac{2\overrightarrow{AC}}{3+2} = \frac{2\vec{b}}{5}$.Thus,$\overrightarrow{BE} = \overrightarrow{AE} - \overrightarrow{AB} = \frac{2\vec{b}}{5} - \vec{c} = \frac{2\vec{b} - 5\vec{c}}{5}$.Since $F$ divides $AB$ in ratio $3:7$,$\overrightarrow{AF} = \frac{3\overrightarrow{AB}}{3+7} = \frac{3\vec{c}}{10}$.Thus,$\overrightarrow{CF} = \overrightarrow{AF} - \overrightarrow{AC} = \frac{3\vec{c}}{10} - \vec{b} = \frac{3\vec{c} - 10\vec{b}}{10}$.Now,$\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = \frac{4\vec{c} + \vec{b}}{5} + \frac{2\vec{b} - 5\vec{c}}{5} + \frac{3\vec{c} - 10\vec{b}}{10} = \frac{8\vec{c} + 2\vec{b} + 4\vec{b} - 10\vec{c} + 3\vec{c} - 10\vec{b}}{10} = \frac{\vec{c} - 4\vec{b}}{10}$.Point $K$ divides $AB$ in ratio $1:3$,so $\overrightarrow{AK} = \frac{1}{4}\overrightarrow{AB} = \frac{\vec{c}}{4}$.Then $\overrightarrow{CK} = \overrightarrow{AK} - \overrightarrow{AC} = \frac{\vec{c}}{4} - \vec{b} = \frac{\vec{c} - 4\vec{b}}{4}$.Ratio = $\frac{(\vec{c} - 4\vec{b})/10}{(\vec{c} - 4\vec{b})/4} = \frac{4}{10} = \frac{2}{5}$.

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

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