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

Give the equations to find the magnitude and direction of the resultant vector of two vectors $\vec{A}$ and $\vec{B}$ that are inclined at an angle $\theta$ to each other.

If $\sqrt{A^2+B^2}$ represents the magnitude of the resultant of two vectors $(\vec{A}+\vec{B})$ and $(\vec{A}-\vec{B})$,then the angle between the two vectors $\vec{A}$ and $\vec{B}$ is

When $n$ vectors of different magnitudes are added,we get a null vector. Then the value of $n$ cannot be

If $r_1=2 \hat{x}$ and $r_2=2 \hat{y}$, where $\hat{x}$ and $\hat{y}$ are unit vectors along the $X$-axis and $Y$-axis respectively, then the magnitude of $r_1+r_2$ is

Statement $I:$ Two forces $(\overrightarrow{P}+\overrightarrow{Q})$ and $(\overrightarrow{P}-\overrightarrow{Q})$,where $\overrightarrow{P} \perp \overrightarrow{Q}$,act at an angle $\theta_{1}$ to each other,and the magnitude of their resultant is $\sqrt{3(P^{2}+Q^{2})}$. When they act at an angle $\theta_{2}$,the magnitude of their resultant becomes $\sqrt{2(P^{2}+Q^{2})}$. This is possible only when $\theta_{1} < \theta_{2}$.
Statement $II:$ In the situation given above,$\theta_{1} = 60^{\circ}$ and $\theta_{2} = 90^{\circ}$.
In the light of the above statements,choose the most appropriate answer from the options given below.