In a quadrilateral PQRS,M and N are mid-point | vedclass

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(D) Let $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{m}, \vec{n}$ be the position vectors of points $P, Q, R, S, M, N$ respectively.We have $\vec{PS} + \

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$2$

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(D) Let $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{m}, \vec{n}$ be the position vectors of points $P, Q, R, S, M, N$ respectively.We have $\vec{PS} + \vec{QR} = (\vec{s} - \vec{p}) + (\vec{r} - \vec{q}) = (\vec{s} + \vec{r}) - (\vec{p} + \vec{q})$.Since $M$ is the mid-point of $PQ$,$\vec{m} = \frac{\vec{p} + \vec{q}}{2}$,which implies $\vec{p} + \vec{q} = 2\vec{m}$.Since $N$ is the mid-point of $RS$,$\vec{n} = \frac{\vec{r} + \vec{s}}{2}$,which implies $\vec{r} + \vec{s} = 2\vec{n}$.Substituting these into the expression:$\vec{PS} + \vec{QR} = 2\vec{n} - 2\vec{m} = 2(\vec{n} - \vec{m}) = 2\vec{MN}$.Comparing this with $\vec{PS} + \vec{QR} = t \vec{MN}$,we get $t = 2$.

In a quadrilateral $PQRS$,$M$ and $N$ are mid-points of the sides $PQ$ and $RS$ respectively. If $\vec{PS} + \vec{QR} = t \vec{MN}$,then $t =$

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