In a triangle ABC, D and E divide the sides B | vedclass

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(B) Let $A, B, C, D, E, P$ have position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{e}, \vec{p}$ respectively.$D$ divides $BC$ in the ratio $2:

Vedclass · Accepted Answer

$3:4$

Vedclass · Answer

(B) Let $A, B, C, D, E, P$ have position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{e}, \vec{p}$ respectively.$D$ divides $BC$ in the ratio $2:1$, so $\vec{d} = \frac{2\vec{c} + \vec{b}}{3} \Rightarrow 3\vec{d} = 2\vec{c} + \vec{b} \quad (1)$.$E$ divides $CA$ in the ratio $2:1$, so $\vec{e} = \frac{2\vec{a} + \vec{c}}{3} \Rightarrow 3\vec{e} = 2\vec{a} + \vec{c} \quad (2)$.From $(1)$, $2\vec{c} = 3\vec{d} - \vec{b}$.From $(2)$, $\vec{c} = 3\vec{e} - 2\vec{a} \Rightarrow 2\vec{c} = 6\vec{e} - 4\vec{a}$.Equating the two expressions for $2\vec{c}$:$3\vec{d} - \vec{b} = 6\vec{e} - 4\vec{a} \Rightarrow 4\vec{a} + 3\vec{d} = 6\vec{e} + \vec{b}$.Dividing by $7$:$\frac{4\vec{a} + 3\vec{d}}{7} = \frac{6\vec{e} + \vec{b}}{7}$.This point $\vec{p}$ divides $AD$ in the ratio $3:4$ and $BE$ in the ratio $1:6$. However, the question asks for the ratio in which $P$ divides $AD$. Based on the section formula $\frac{m\vec{x_2} + n\vec{x_1}}{m+n}$, the ratio is $3:4$.

In a triangle $ABC$, $D$ and $E$ divide the sides $BC$ and $CA$ in the ratio $2:1$ respectively. If $P$ is the point of intersection of $AD$ and $BE$, then the ratio in which $P$ divides $AD$ is

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