Given below in Column $-I$ are the relations between vectors $\vec a$,$\vec b$,and $\vec c$,and in Column $-II$ are the orientations of $\vec a$,$\vec b$,and $\vec c$ in the $XY-$ plane. Match the relation in Column $-I$ to the correct orientations in Column $-II$.

Column $-I$	Column $-II$
$(a) \vec a + \vec b = \vec c$	$(i)$ Vector $\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ connects the origin to the tip of $\vec c$
$(b) \vec a - \vec c = \vec b$	$(ii)$ Vector $\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ connects the origin to the tip of $\vec b$
$(c) \vec b - \vec a = \vec c$	$(iii)$ Vector $\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$
$(d) \vec a + \vec b + \vec c = 0$	$(iv)$ Vector $\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ connects the origin to the tip of $\vec b$

(A-II, B-III, C-I, D-IV) Using the triangle law of vector addition,where the resultant vector is the one that closes the triangle in the opposite direction of the other two vectors connected head-to-tail.
$(a)$ For $\vec a + \vec b = \vec c$,the vectors $\vec a$ and $\vec b$ must be connected head-to-tail,and $\vec c$ is the resultant. In diagram $(ii)$,$\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ is the resultant. Thus,$(a) \rightarrow (ii)$.
$(b)$ For $\vec a - \vec c = \vec b$,we can write $\vec a = \vec b + \vec c$. In diagram $(iii)$,$\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$. Thus,$\vec c + \vec b = \vec a$,which implies $\vec a - \vec c = \vec b$. Thus,$(b) \rightarrow (iii)$.
$(c)$ For $\vec b - \vec a = \vec c$,we can write $\vec b = \vec a + \vec c$. In diagram $(i)$,$\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ is the resultant. Thus,$(c) \rightarrow (i)$.
$(d)$ For $\vec a + \vec b + \vec c = 0$,the vectors must form a closed loop. In diagram $(iv)$,$\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ is the vector closing the loop. Thus,$(d) \rightarrow (iv)$.