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(C) Given $\overrightarrow{OA} = a$ and $\overrightarrow{OB} = b$.Point $C$ lies on $OA$ such that $2AC = CO$. This implies $C$ divides $OA$ in the ra

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$3b - \frac{a}{3}$

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(C) Given $\overrightarrow{OA} = a$ and $\overrightarrow{OB} = b$.Point $C$ lies on $OA$ such that $2AC = CO$. This implies $C$ divides $OA$ in the ratio $1:2$ from $O$ to $A$.Using the section formula,$\overrightarrow{OC} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}a$.Since $CD$ is parallel to $OB$ and $|\overrightarrow{CD}| = 3|\overrightarrow{OB}|$,we have $\overrightarrow{CD} = 3\overrightarrow{OB} = 3b$.Now,$\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD} = \frac{2}{3}a + 3b$.Finally,$\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = (\frac{2}{3}a + 3b) - a = 3b - \frac{1}{3}a$.

Let $A$ and $B$ be points with position vectors $a$ and $b$ with respect to the origin $O$. If the point $C$ on $OA$ is such that $2AC = CO$,$CD$ is parallel to $OB$ and $|\overrightarrow{CD}| = 3|\overrightarrow{OB}|$,then $\overrightarrow{AD}$ is equal to

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