In triangle PQR,M is the mid-point of QR and | vedclass

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(D) Let the position vectors of $P, Q, R$ be $\vec{p}, \vec{q}, \vec{r}$ respectively. Since $M$ is the mid-point of $QR$,the position vector of $M$ i

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$4$

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(D) Let the position vectors of $P, Q, R$ be $\vec{p}, \vec{q}, \vec{r}$ respectively. Since $M$ is the mid-point of $QR$,the position vector of $M$ is $\vec{m} = \frac{\vec{q} + \vec{r}}{2}$. Since $C$ is the mid-point of $PM$,the position vector of $C$ is $\vec{c} = \frac{\vec{p} + \vec{m}}{2} = \frac{\vec{p} + \frac{\vec{q} + \vec{r}}{2}}{2} = \frac{2\vec{p} + \vec{q} + \vec{r}}{4}$. The line $QC$ passes through $Q$ and $C$. The vector equation of line $QC$ is $\vec{r} = \vec{q} + t(\vec{c} - \vec{q}) = \vec{q} + t(\frac{2\vec{p} + \vec{q} + \vec{r}}{4} - \vec{q}) = \vec{q} + t(\frac{2\vec{p} - 3\vec{q} + \vec{r}}{4})$. The line $PR$ passes through $P$ and $R$. The vector equation of line $PR$ is $\vec{r} = \vec{p} + s(\vec{r} - \vec{p}) = (1-s)\vec{p} + s\vec{r}$. Equating the coefficients of $\vec{p}, \vec{q}, \vec{r}$ for the intersection point $N$,we find $t = \frac{4}{3}$ and $s = \frac{1}{3}$. Thus,$\vec{n} = (1 - \frac{1}{3})\vec{p} + \frac{1}{3}\vec{r} = \frac{2\vec{p} + \vec{r}}{3}$. Since $\vec{n} = \vec{q} + \frac{4}{3}(\frac{2\vec{p} - 3\vec{q} + \vec{r}}{4}) = \vec{q} + \frac{2\vec{p} - 3\vec{q} + \vec{r}}{3} = \frac{2\vec{p} + \vec{r}}{3}$,the point $N$ lies on $PR$. Using the section formula,$N$ divides $QC$ in the ratio $t : (1-t) = \frac{4}{3} : (1 - \frac{4}{3}) = \frac{4}{3} : -\frac{1}{3} = 4 : -1$. Thus,$\overrightarrow{QN} = 4\overrightarrow{QC}$,which implies $\frac{|\overrightarrow{QN}|}{|\overrightarrow{CN}|} = 4$.

In $\triangle PQR$,$M$ is the mid-point of $QR$ and $C$ is the mid-point of $PM$. If $QC$ when extended meets $PR$ at $N$,then $\frac{|\overrightarrow{QN}|}{|\overrightarrow{CN}|}=$

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