D, E, F are points on the sides BC, CA and AB | vedclass

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(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Using the section formula,the position vectors of $D,

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$\frac{5}{6}$

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(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Using the section formula,the position vectors of $D, E, F$ are:$\vec{d} = \frac{3\vec{b} + 2\vec{c}}{5}$,$\vec{e} = \frac{2\vec{c} + 1\vec{a}}{3}$,$\vec{f} = \frac{1\vec{a} + 3\vec{b}}{4}$.Any point $P$ on $AD$ can be written as $\vec{p} = (1-t)\vec{a} + t\vec{d} = (1-t)\vec{a} + t(\frac{3\vec{b} + 2\vec{c}}{5}) = (1-t)\vec{a} + \frac{3t}{5}\vec{b} + \frac{2t}{5}\vec{c}$.Since $P$ also lies on $BE$,$\vec{p} = (1-m)\vec{b} + m\vec{e} = (1-m)\vec{b} + m(\frac{2\vec{c} + \vec{a}}{3}) = \frac{m}{3}\vec{a} + (1-m)\vec{b} + \frac{2m}{3}\vec{c}$.Comparing the coefficients of $\vec{a}, \vec{b}, \vec{c}$ (since $\vec{a}, \vec{b}, \vec{c}$ are linearly independent in the plane):$1-t = \frac{m}{3}$,$\frac{3t}{5} = 1-m$,$\frac{2t}{5} = \frac{2m}{3}$.From $\frac{2t}{5} = \frac{2m}{3}$,we get $3t = 5m$,so $m = \frac{3t}{5}$.Substituting into $1-t = \frac{m}{3}$,we get $1-t = \frac{3t}{15} = \frac{t}{5}$,so $1 = \frac{6t}{5}$,which gives $t = \frac{5}{6}$.Then $m = \frac{3}{5} 	imes \frac{5}{6} = \frac{1}{2}$.Substituting $t = \frac{5}{6}$ into the expression for $\vec{p}$:$\vec{p} = (1-\frac{5}{6})\vec{a} + \frac{3}{5}(\frac{5}{6})\vec{b} + \frac{2}{5}(\frac{5}{6})\vec{c} = \frac{1}{6}\vec{a} + \frac{1}{2}\vec{b} + \frac{1}{3}\vec{c}$.Now,$\overrightarrow{AP} = \vec{p} - \vec{a} = \frac{1}{6}\vec{a} + \frac{1}{2}\vec{b} + \frac{1}{3}\vec{c} - \vec{a} = \frac{1}{2}\vec{b} + \frac{1}{3}\vec{c} - \frac{5}{6}\vec{a}$.Since $\vec{a} + \vec{b} + \vec{c}$ is not the correct approach here,we use $\vec{a} + \vec{b} + \vec{c}$ is not needed,rather $\overrightarrow{AP} = \frac{1}{2}(\vec{b}-\vec{a}) + \frac{1}{3}(\vec{c}-\vec{a}) = \frac{1}{2}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC}$.Thus,$x_1 = \frac{1}{2}$ and $y_1 = \frac{1}{3}$.Therefore,$x_1 + y_1 = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.

$D, E, F$ are points on the sides $BC, CA$ and $AB$ of a $\triangle ABC$ respectively,dividing them in the ratios $2:3, 1:2, 3:1$ internally. The lines $BE$ and $CF$ intersect on the line $AD$ at $P$. If $\overrightarrow{AP} = x_1 \overrightarrow{AB} + y_1 \overrightarrow{AC}$,then $x_1 + y_1 =$

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