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REPEATER 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20244 Marks
Question
In $\triangle A B C, D$ and $E$ are points on $B C$ and $A C$ respectively, such that $B D=2 D C$ and $A E=$
$3 E C$. Let $P$ be the point of intersection of $A D$ and $B E$. Find $\frac{B P}{P E}$ using vector method.
✓

Answer

Let $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $A, B$ and $C$ respectively with respect to some origin.
Image
$D$ divides $B C$ in the ratio $2:1$ and $E$ divides $A C$ in the ratio $3: 1$.
$\therefore \bar{d}=\frac{\bar{b}+2 c}{3} $ and $ \bar{e}=\frac{\bar{a}+3 c}{4}$
Let point of intersection $P$ of $A D$ and $B E$ divides $B E$ in the ratio $k: 1$ and $A D$ in the ratio $m: 1,$ then position vectors of $P$ in these two cases are
$\frac{\bar{b}+k\left(\frac{\bar{a}+3 \bar{c}}{4}\right)}{k+1} \text { and } \frac{\bar{a}+m\left(\frac{b+2 \bar{c}}{3}\right)}{m+1}$ respectively.
Equating the position vectors of $P$ we get,
$\frac{k}{4(k+1)} \bar{a}+\frac{1}{k+1} \bar{b}+\frac{3 k}{4(k+1)} \bar{c}$
$=\frac{1}{m+1} \bar{a}+\frac{m}{3(m+1)} \bar{b}+\frac{2 m}{3(m+1)} \bar{c}$
$\therefore \frac{k}{4(k+1)}=\frac{1}{m+1}.......(1)$
$\therefore \frac{1}{k+1}=\frac{m}{3(m+1)}......(2)$
$\therefore \frac{3 k}{4(k+1)}=\frac{2 m}{3(m+1)}......(3)$
Dividing equation $(3)$ by equation $(2),$
$\frac{\frac{3 k}{4(k+1)}}{\frac{1}{k+1}}$
$=\frac{\frac{2 m}{3(m+1)}}{\frac{m}{3(m+1)}}$
$\frac{3 k}{4}=2$
$k=\frac{8}{3}$
$\therefore \frac{B P}{P E}=\frac{k}{1}$
$\therefore \frac{B P}{P E}=\frac{8}{3}$

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