Five points given by A, B, C, D, E are in a p | vedclass

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(B) The resultant force $R$ is the sum of all the given vectors:$R = (\overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE}) + (\overrightar

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$3\overrightarrow{AB}$

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(B) The resultant force $R$ is the sum of all the given vectors:$R = (\overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE}) + (\overrightarrow{CB} + \overrightarrow{DB} + \overrightarrow{EB})$By rearranging the terms using the commutative property of vector addition:$R = (\overrightarrow{AC} + \overrightarrow{CB}) + (\overrightarrow{AD} + \overrightarrow{DB}) + (\overrightarrow{AE} + \overrightarrow{EB})$Using the triangle law of vector addition,where $\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$:$\overrightarrow{AC} + \overrightarrow{CB} = \overrightarrow{AB}$$\overrightarrow{AD} + \overrightarrow{DB} = \overrightarrow{AB}$$\overrightarrow{AE} + \overrightarrow{EB} = \overrightarrow{AB}$Substituting these into the expression for $R$:$R = \overrightarrow{AB} + \overrightarrow{AB} + \overrightarrow{AB} = 3\overrightarrow{AB}$.

Five points given by $A, B, C, D, E$ are in a plane. Three forces $\overrightarrow{AC}, \overrightarrow{AD},$ and $\overrightarrow{AE}$ act at $A$,and three forces $\overrightarrow{CB}, \overrightarrow{DB},$ and $\overrightarrow{EB}$ act at $B$. Then their resultant is:

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