Let PQR be a triangle. The points A, B and C | vedclass

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(B) Let the position vectors of vertices $P, Q, R$ be $\vec{p}, \vec{q}, \vec{r}$ respectively. Without loss of generality,let $\vec{p} = \vec{0}$.Giv

Vedclass · Accepted Answer

$3$

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(B) Let the position vectors of vertices $P, Q, R$ be $\vec{p}, \vec{q}, \vec{r}$ respectively. Without loss of generality,let $\vec{p} = \vec{0}$.Given $\frac{QA}{AR} = \frac{1}{2}$,point $A$ divides $QR$ in ratio $1:2$. Thus,$\vec{a} = \frac{2\vec{q} + 1\vec{r}}{3}$.Given $\frac{RB}{BP} = \frac{1}{2}$,point $B$ divides $RP$ in ratio $1:2$. Thus,$\vec{b} = \frac{2\vec{r} + 1\vec{p}}{3} = \frac{2\vec{r}}{3}$.Given $\frac{PC}{CQ} = \frac{1}{2}$,point $C$ divides $PQ$ in ratio $1:2$. Thus,$\vec{c} = \frac{2\vec{p} + 1\vec{q}}{3} = \frac{\vec{q}}{3}$.The area of $	riangle PQR$ is given by $\Delta = \frac{1}{2} |\vec{q} 	imes \vec{r}|$.The area of $	riangle ABC$ is given by $\Delta' = \frac{1}{2} |(\vec{b} - \vec{a}) 	imes (\vec{c} - \vec{a})|$.Calculating vectors: $\vec{b} - \vec{a} = \frac{2\vec{r} - 2\vec{q} - \vec{r}}{3} = \frac{\vec{r} - 2\vec{q}}{3}$ and $\vec{c} - \vec{a} = \frac{\vec{q} - 2\vec{q} - \vec{r}}{3} = \frac{-\vec{q} - \vec{r}}{3}$.$\Delta' = \frac{1}{2} |\frac{1}{9} (\vec{r} - 2\vec{q}) 	imes (-\vec{q} - \vec{r})| = \frac{1}{18} |-\vec{r} 	imes \vec{q} - \vec{r} 	imes \vec{r} + 2\vec{q} 	imes \vec{q} + 2\vec{q} 	imes \vec{r}|$.Since $\vec{x} 	imes \vec{x} = 0$ and $\vec{r} 	imes \vec{q} = -\vec{q} 	imes \vec{r}$,we have $\Delta' = \frac{1}{18} |\vec{q} 	imes \vec{r} + 2\vec{q} 	imes \vec{r}| = \frac{1}{18} |3(\vec{q} 	imes \vec{r})| = \frac{1}{6} |\vec{q} 	imes \vec{r}|$.Thus,$\frac{\operatorname{Area}(	riangle PQR)}{\operatorname{Area}(	riangle ABC)} = \frac{\frac{1}{2} |\vec{q} 	imes \vec{r}|}{\frac{1}{6} |\vec{q} 	imes \vec{r}|} = 3$.

Let $PQR$ be a triangle. The points $A, B$ and $C$ are on the sides $QR, RP$ and $PQ$ respectively such that $\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$. Then $\frac{\operatorname{Area}(\triangle PQR)}{\operatorname{Area}(\triangle ABC)}$ is equal to $........$

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