The resultant of two forces acting at an angle of $120^{\circ}$ is $10 \,kg$-wt and is perpendicular to one of the forces. That force is

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 $\vec b$

Three vectors each of magnitude $3 \sqrt{1.5}$ units are acting at a point. If the angle between any two vectors is $\frac{\pi}{3}$,then the magnitude of the resultant vector of the three vectors is

Two equal forces ($P$ each) act at a point inclined to each other at an angle of $120^\circ$. The magnitude of their resultant is

$A$ particle moves $21 \ m$ along the direction of the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$,then it moves $14 \ m$ along the direction of the vector $3\hat{i} - 2\hat{j} + 6\hat{k}$. What is its total displacement (in meters)?

The magnitudes of pairs of displacement vectors are given. Which pair of displacement vectors cannot be added to give a resultant vector of magnitude $13 \, cm$?