$A$ force ${F_1} = 1\,N$ acts along the direction of $x = 0$,and a force ${F_2} = 2\,N$ acts along the direction of $y = 0$. Find the resultant force.

Resultant of two vectors $\vec{P}$ and $\vec{Q}$ is of magnitude $A$. If $\vec{Q}$ is reversed, then the resultant is of magnitude $B$. The value of $A^2 + B^2$ is

Vectors $\vec{A}$ and $\vec{B}$ make angles of $20^\circ$ and $110^\circ$ with the $x$-axis,respectively. The magnitudes of these vectors are $5 \ m$ and $12 \ m$,respectively. The angle that their resultant vector makes with the $x$-axis is:

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$. What is the angle between $\vec{A}$ and $\vec{B}$?

What displacement must be added to the displacement $25\hat{i} - 6\hat{j} \text{ m}$ to give a resultant displacement of $7.0 \text{ m}$ pointing in the $X$-direction?

Two vectors $\vec{A}$ and $\vec{B}$ are such that $\vec{A} + \vec{B} = \vec{A} - \vec{B}$. Then: