A point C with position vector frac{3 bar{a}+ | vedclass

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(D) Let the position vectors of points $A$ and $B$ be $\vec{r}_A$ and $\vec{r}_B$ respectively.Given $\vec{r}_A = \bar{a}-2 \bar{b}+3 \bar{c}$.Point $

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$\bar{a}+3 \bar{b}-4 \bar{c}$

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(D) Let the position vectors of points $A$ and $B$ be $\vec{r}_A$ and $\vec{r}_B$ respectively.Given $\vec{r}_A = \bar{a}-2 \bar{b}+3 \bar{c}$.Point $C$ divides the line segment $AB$ in the ratio $m:n = 2:1$.The position vector of $C$ is given by the section formula: $\vec{r}_C = \frac{m \vec{r}_B + n \vec{r}_A}{m+n}$.Substituting the given values: $\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3} = \frac{2 \vec{r}_B + 1 (\bar{a}-2 \bar{b}+3 \bar{c})}{2+1}$.$\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3} = \frac{2 \vec{r}_B + \bar{a}-2 \bar{b}+3 \bar{c}}{3}$.Equating the numerators: $3 \bar{a}+4 \bar{b}-5 \bar{c} = 2 \vec{r}_B + \bar{a}-2 \bar{b}+3 \bar{c}$.$2 \vec{r}_B = (3 \bar{a}-\bar{a}) + (4 \bar{b}+2 \bar{b}) + (-5 \bar{c}-3 \bar{c})$.$2 \vec{r}_B = 2 \bar{a} + 6 \bar{b} - 8 \bar{c}$.$\vec{r}_B = \bar{a} + 3 \bar{b} - 4 \bar{c}$.Thus,the position vector of $B$ is $\bar{a} + 3 \bar{b} - 4 \bar{c}$.

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