OABCD is a pentagon in which the sides OA and | vedclass

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(D) Given,$\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$.Since $OA \parallel CB$ and $\frac{OA}{CB}=2$,we have $\vec{CB} = \frac{1}{2}\vec{OA} = \frac{1}{2}\vec

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$6\vec{d}$

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(D) Given,$\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$.Since $OA \parallel CB$ and $\frac{OA}{CB}=2$,we have $\vec{CB} = \frac{1}{2}\vec{OA} = \frac{1}{2}\vec{a}$.Since $OD \parallel AB$ and $\frac{OD}{AB}=\frac{1}{3}$,we have $\vec{AB} = 3\vec{OD} = 3\vec{d}$.Now,we express the required sum in terms of vectors $\vec{a}$ and $\vec{d}$:$\vec{AD} = \vec{OD} - \vec{OA} = \vec{d} - \vec{a}$.$\vec{OC} = \vec{OD} + \vec{DC}$.From the pentagon geometry,$\vec{OC} = \vec{OA} + \vec{AB} + \vec{BC} = \vec{a} + 3\vec{d} - \frac{1}{2}\vec{a} = \frac{1}{2}\vec{a} + 3\vec{d}$.Also,$\vec{DC} = \vec{OC} - \vec{OD} = (\frac{1}{2}\vec{a} + 3\vec{d}) - \vec{d} = \frac{1}{2}\vec{a} + 2\vec{d}$.Summing these: $\vec{AD} + \vec{OC} + \vec{DC} = (\vec{d} - \vec{a}) + (\frac{1}{2}\vec{a} + 3\vec{d}) + (\frac{1}{2}\vec{a} + 2\vec{d})$.$= (-\vec{a} + \frac{1}{2}\vec{a} + \frac{1}{2}\vec{a}) + (\vec{d} + 3\vec{d} + 2\vec{d}) = 0\vec{a} + 6\vec{d} = 6\vec{d}$.

$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. Also,it is given that $\frac{OA}{CB}=2$,$\frac{OD}{AB}=\frac{1}{3}$. If $\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$,then $\vec{AD}+\vec{OC}+\vec{DC}=$

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