If PQRST is a pentagon,then the resultant of | vedclass

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(C) Let $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{t}$ be the position vectors of the vertices $P, Q, R, S, T$ respectively. The given sum of vectors i

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$3 \overline{PS}$

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(C) Let $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{t}$ be the position vectors of the vertices $P, Q, R, S, T$ respectively. The given sum of vectors is $\vec{V} = \overline{PQ} + \overline{PT} + \overline{QR} + \overline{SR} + \overline{TS} + \overline{PS}$. Substituting the position vectors: $\vec{V} = (\vec{q} - \vec{p}) + (\vec{t} - \vec{p}) + (\vec{r} - \vec{q}) + (\vec{s} - \vec{r}) + (\vec{s} - \vec{t}) + (\vec{s} - \vec{p})$. Grouping the terms: $\vec{V} = (\vec{q} - \vec{q}) + (\vec{t} - \vec{t}) + (\vec{r} - \vec{r}) + (\vec{s} + \vec{s} + \vec{s}) - (\vec{p} + \vec{p} + \vec{p})$. $\vec{V} = 3\vec{s} - 3\vec{p} = 3(\vec{s} - \vec{p})$. Since $\vec{s} - \vec{p} = \overline{PS}$,we have $\vec{V} = 3\overline{PS}$.

If $PQRST$ is a pentagon,then the resultant of forces $\overline{PQ}, \overline{PT}, \overline{QR}, \overline{SR}, \overline{TS}$ and $\overline{PS}$ is

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