AB=a and AC=b are the sides of triangle ABC. | vedclass

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(A) Let the position vectors of vertices $A, B, C$ be $\vec{0}, \vec{a}, \vec{b}$ respectively. Since $P$ lies on $AB$ such that $AP:PB = 1:2$,the pos

Vedclass · Accepted Answer

$\frac{49}{4}$

Vedclass · Answer

(A) Let the position vectors of vertices $A, B, C$ be $\vec{0}, \vec{a}, \vec{b}$ respectively. Since $P$ lies on $AB$ such that $AP:PB = 1:2$,the position vector of $P$ is $\vec{p} = \frac{1}{3}\vec{a}$. Since $Q$ lies on $BC$ such that $BQ:QC = 1:2$,the position vector of $Q$ is $\vec{q} = \frac{2\vec{b} + \vec{a}}{3}$. Let $D$ be the intersection of $AQ$ and $CP$. $D$ lies on $AQ$,so $\vec{d} = (1-t)\vec{A} + t\vec{Q} = t(\frac{2\vec{b} + \vec{a}}{3}) = \frac{t}{3}\vec{a} + \frac{2t}{3}\vec{b}$. $D$ lies on $CP$,so $\vec{d} = (1-s)\vec{C} + s\vec{P} = (1-s)\vec{b} + s(\frac{1}{3}\vec{a}) = \frac{s}{3}\vec{a} + (1-s)\vec{b}$. Comparing coefficients of $\vec{a}$ and $\vec{b}$: $\frac{t}{3} = \frac{s}{3} \Rightarrow t = s$ $\frac{2t}{3} = 1 - s \Rightarrow \frac{2s}{3} + s = 1 \Rightarrow \frac{5s}{3} = 1 \Rightarrow s = \frac{3}{5}$. Thus,$\vec{d} = \frac{1}{5}\vec{a} + \frac{2}{5}\vec{b}$. The area of $	riangle ABC = \frac{1}{2}|\vec{a} 	imes \vec{b}|$. The area of $	riangle BCD = \frac{1}{2}|(\vec{c}-\vec{b}) 	imes (\vec{d}-\vec{b})| = \frac{1}{2}|(\vec{b}-\vec{b}) 	imes (\frac{1}{5}\vec{a} + \frac{2}{5}\vec{b} - \vec{b})| = \frac{1}{2}|\vec{b} 	imes (\frac{1}{5}\vec{a} - \frac{3}{5}\vec{b})| = \frac{1}{2}|\frac{1}{5}(\vec{b} 	imes \vec{a})| = \frac{1}{10}|\vec{a} 	imes \vec{b}|$. Given area of $	riangle BCD = 7$,so $\frac{1}{10}|\vec{a} 	imes \vec{b}| = 7 \Rightarrow \frac{1}{2}|\vec{a} 	imes \vec{b}| = 35$. Wait,re-evaluating the ratio: $Area(	riangle BCD) = \frac{1}{5} Area(	riangle ABC)$ is incorrect. Let's use mass point geometry. Assign mass $1$ at $C$. Since $BQ:QC=1:2$,mass at $B$ is $2$. Since $AP:PB=1:2$,mass at $A$ is $4$. Total mass at $P$ is $1+4=5$. Total mass at $Q$ is $1+2=3$. The intersection $D$ has mass $4+1+2=7$. Area ratio $\frac{Area(	riangle BCD)}{Area(	riangle ABC)} = \frac{Mass(A)}{Mass(A)+Mass(B)+Mass(C)} = \frac{4}{4+2+1} = \frac{4}{7}$. Given $Area(	riangle BCD) = 7$,then $Area(	riangle ABC) = 7 	imes \frac{7}{4} = \frac{49}{4}$. Thus,option $(a)$ is correct.

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