Suppose ABCDE is a pentagon. The resultant ve | vedclass

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(A) We need to find the sum of the vectors: $\vec{S} = \vec{AB} + \vec{AE} + \vec{BC} + \vec{DC} + \vec{ED} + \vec{AC}$.By rearranging the terms, we g

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$3 \vec{AC}$

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(A) We need to find the sum of the vectors: $\vec{S} = \vec{AB} + \vec{AE} + \vec{BC} + \vec{DC} + \vec{ED} + \vec{AC}$.By rearranging the terms, we get:$\vec{S} = (\vec{AB} + \vec{BC}) + (\vec{AE} + \vec{ED}) + (\vec{DC} + \vec{AC})$.Using the triangle law of vector addition, we know that $\vec{AB} + \vec{BC} = \vec{AC}$ and $\vec{AE} + \vec{ED} = \vec{AD}$.Substituting these into the expression, we get:$\vec{S} = \vec{AC} + \vec{AD} + (\vec{DC} + \vec{AC})$.Since $\vec{AD} + \vec{DC} = \vec{AC}$ (by the triangle law of vector addition in $	riangle ADC$),$\vec{S} = \vec{AC} + (\vec{AD} + \vec{DC}) + \vec{AC} = \vec{AC} + \vec{AC} + \vec{AC} = 3 \vec{AC}$.

Suppose $ABCDE$ is a pentagon. The resultant vector of the vectors $\vec{AB}, \vec{AE}, \vec{BC}, \vec{DC}, \vec{ED}$ and $\vec{AC}$ is

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