Add vectors $\overrightarrow{ A }, \overrightarrow{ B }$ and $\overrightarrow{ C }$ each having magnitude of $50$ unit and inclined to the $X$-axis at angles $45^{\circ}, 135^{\circ}$ and $315^{\circ}$ respectively.

The magnitude of vectors $\overrightarrow{ OA }, \overrightarrow{ OB }$ and $\overrightarrow{ OC }$ in the given figure are equal. The direction of $\overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }$ with $x$-axis will be

The vector $\overrightarrow{O A}$ where $O$ is origin is given by $\overrightarrow{O A}=2 \hat{i}+2 \hat{j}$. Now it is rotated by $45^{\circ}$ anticlockwise about $O$. What will be the new vector?

Two forces are such that the sum of their magnitudes is $18\; N$ and their resultant is $12\; N$ which is perpendicular to the smaller force. Then the magnitudes of the forces are

Assertion $A$ : If $A, B, C, D$ are four points on a semi-circular arc with centre at $'O'$ such that $|\overrightarrow{{AB}}|=|\overrightarrow{{BC}}|=|\overrightarrow{{CD}}|$, then $\overrightarrow{{AB}}+\overrightarrow{{AC}}+\overrightarrow{{AD}}=4 \overrightarrow{{AO}}+\overrightarrow{{OB}}+\overrightarrow{{OC}}$

Reason $R$ : Polygon law of vector addition yields $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{A D}=2 \overrightarrow{A O}$

In the light of the above statements, choose the most appropriate answer from the options given below