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(A) Step $1$: Applying the triangle law of vector addition.In $	riangle OPR$,by the triangle law of vector addition:$\vec{a} = \vec{c} + \overrightar

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$\vec{a} + \vec{b} = 2\vec{c}$

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(A) Step $1$: Applying the triangle law of vector addition.In $	riangle OPR$,by the triangle law of vector addition:$\vec{a} = \vec{c} + \overrightarrow{RP} \quad (1)$In $	riangle ORQ$,by the triangle law of vector addition:$\vec{b} = \overrightarrow{RQ} - \vec{c} \implies \vec{c} + \vec{b} = \overrightarrow{RQ} \implies \vec{b} = \overrightarrow{RQ} - \vec{c}$ (Wait,let's re-evaluate the geometry).Correct approach: In $	riangle OPR$,$\vec{a} = \vec{c} + \overrightarrow{RP}$.In $	riangle ORQ$,$\vec{c} = \vec{b} + \overrightarrow{RQ} \implies \overrightarrow{RQ} = \vec{c} - \vec{b}$.Since $R$ is the midpoint of $PQ$,$\overrightarrow{PR} + \overrightarrow{RQ} = 0$,so $\overrightarrow{RP} = -\overrightarrow{RQ}$.From $	riangle OPR$,$\overrightarrow{RP} = \vec{a} - \vec{c}$.From $	riangle ORQ$,$\overrightarrow{RQ} = \vec{c} - \vec{b}$.Since $\overrightarrow{RP} + \overrightarrow{RQ} = 0$,we have $(\vec{a} - \vec{c}) + (\vec{c} - \vec{b}) = 0$ is incorrect. Let's use position vectors.Let origin be $O$. Then $\vec{a} = \vec{P}$,$\vec{b} = \vec{Q}$,$\vec{c} = \vec{R}$.Since $R$ is the midpoint of $PQ$,$\vec{R} = \frac{\vec{P} + \vec{Q}}{2}$.Therefore,$\vec{c} = \frac{\vec{a} + \vec{b}}{2}$,which implies $\vec{a} + \vec{b} = 2\vec{c}$.

The figure shows three vectors $\vec{a}, \vec{b}$ and $\vec{c}$,where $R$ is the midpoint of $PQ$. Which of the following relations is correct?

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