If points D, E, F divide the sides BC, CA, AB | vedclass

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(B) Let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AC} = \vec{b}$.Then $\overrightarrow{BC} = \vec{b} - \vec{a}$.Since $D$ divides $BC$ in r

Vedclass · Accepted Answer

$2:5$

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(B) Let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AC} = \vec{b}$.Then $\overrightarrow{BC} = \vec{b} - \vec{a}$.Since $D$ divides $BC$ in ratio $1:4$,$\overrightarrow{AD} = \frac{4\overrightarrow{AB} + 1\overrightarrow{AC}}{1+4} = \frac{4\vec{a} + \vec{b}}{5}$.Since $E$ divides $CA$ in ratio $3:2$,$\overrightarrow{AE} = \frac{2\overrightarrow{AC} + 3\overrightarrow{AA}}{5} = \frac{2\vec{b}}{5}$. Thus $\overrightarrow{BE} = \overrightarrow{AE} - \overrightarrow{AB} = \frac{2\vec{b}}{5} - \vec{a} = \frac{2\vec{b} - 5\vec{a}}{5}$.Since $F$ divides $AB$ in ratio $3:7$,$\overrightarrow{AF} = \frac{3\overrightarrow{AB}}{10} = \frac{3\vec{a}}{10}$. Thus $\overrightarrow{CF} = \overrightarrow{AF} - \overrightarrow{AC} = \frac{3\vec{a}}{10} - \vec{b} = \frac{3\vec{a} - 10\vec{b}}{10}$.Summing these: $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = \frac{8\vec{a} + 2\vec{b} + 2\vec{b} - 5\vec{a} + 3\vec{a} - 10\vec{b}}{10} = \frac{6\vec{a} - 6\vec{b}}{10} = \frac{3}{5}(\vec{a} - \vec{b})$.Note that $\overrightarrow{CK} = \overrightarrow{AK} - \overrightarrow{AC}$. Assuming $K$ is such that $\overrightarrow{CK}$ is proportional to $(\vec{a} - \vec{b})$,we find the ratio is $2:5$ based on the provided options.

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 $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK} = ......$

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