Let ABC be a triangle and bar{a}, bar{b}, bar | vedclass

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(A) Given that $D$ divides $BC$ in the ratio $2:3$,$E$ divides $AC$ in the ratio $2:1$,and $P$ divides $DE$ in the ratio $3:5$ internally.Using the se

Vedclass · Accepted Answer

$\frac{1}{8}(2 \bar{a}+3 \bar{b}+3 \bar{c})$

Vedclass · Answer

(A) Given that $D$ divides $BC$ in the ratio $2:3$,$E$ divides $AC$ in the ratio $2:1$,and $P$ divides $DE$ in the ratio $3:5$ internally.Using the section formula,the position vectors are:$\vec{d} = \frac{2\vec{c} + 3\vec{b}}{2+3} = \frac{2\vec{c} + 3\vec{b}}{5} \implies 5\vec{d} = 3\vec{b} + 2\vec{c} \quad (i)$$\vec{e} = \frac{2\vec{a} + 1\vec{c}}{2+1} = \frac{2\vec{a} + \vec{c}}{3} \implies 3\vec{e} = 2\vec{a} + \vec{c} \quad (ii)$Now,$P$ divides $DE$ in the ratio $3:5$,so its position vector $\vec{p}$ is:$\vec{p} = \frac{3\vec{e} + 5\vec{d}}{3+5} = \frac{3\vec{e} + 5\vec{d}}{8}$Substitute the expressions for $5\vec{d}$ and $3\vec{e}$ from equations $(i)$ and $(ii)$:$\vec{p} = \frac{(2\vec{a} + \vec{c}) + (3\vec{b} + 2\vec{c})}{8}$$\vec{p} = \frac{2\vec{a} + 3\vec{b} + 3\vec{c}}{8} = \frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$Thus,the correct option is $(A)$.

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