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

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(D) Given the position vectors of $A, B, C$ are $\bar{a}, \bar{b}, \bar{c}$ respectively. Since $D$ divides $BC$ in the ratio $3:1$, the position vect

Vedclass · Accepted Answer

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

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(D) Given the position vectors of $A, B, C$ are $\bar{a}, \bar{b}, \bar{c}$ respectively. Since $D$ divides $BC$ in the ratio $3:1$, the position vector of $D$ is $\bar{d} = \frac{1\bar{b} + 3\bar{c}}{1+3} = \frac{\bar{b} + 3\bar{c}}{4}$. Since $E$ divides $AD$ in the ratio $4:1$, the position vector of $E$ is $\bar{e} = \frac{1\bar{a} + 4\bar{d}}{1+4} = \frac{\bar{a} + 4(\frac{\bar{b} + 3\bar{c}}{4})}{5} = \frac{\bar{a} + \bar{b} + 3\bar{c}}{5}$. Given that $E$ divides $BF$ in the ratio $3:2$, we have $\bar{e} = \frac{2\bar{b} + 3\bar{f}}{2+3} = \frac{2\bar{b} + 3\bar{f}}{5}$. Equating the two expressions for $\bar{e}$: $\frac{\bar{a} + \bar{b} + 3\bar{c}}{5} = \frac{2\bar{b} + 3\bar{f}}{5}$. $\bar{a} + \bar{b} + 3\bar{c} = 2\bar{b} + 3\bar{f}$. $3\bar{f} = \bar{a} - \bar{b} + 3\bar{c}$. $\bar{f} = \frac{\bar{a} - \bar{b} + 3\bar{c}}{3}$.

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