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

The magnitude of the sum of the two vectors $\vec{A}$ and $\vec{B}$ is equal to the magnitude of the difference of the two vectors $\vec{A}$ and $\vec{B}$. The angle between $\vec{A}$ and $\vec{B}$ is: (in $^{\circ}$)

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 magnitude of the resultant vector of these two vectors is ....... $m$.

The resultant $\vec{R}$ of $\vec{P}$ and $\vec{Q}$ is perpendicular to $\vec{P}$. Also,$|\vec{P}| = |\vec{R}|$. Find the angle between $\vec{P}$ and $\vec{Q}$.

If three vectors have equal magnitude,i.e.,$A = B = C$,then the angle between $\vec{A}$ and $\vec{C}$ is $\alpha$. If $\vec{A} + \vec{B} + \vec{C} = 0$,then the angle between $\vec{A}$ and $\vec{C}$ is $\beta$. Find the ratio $\frac{\alpha}{\beta}$.

$A$ 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)?