The resultant of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ has a magnitude $R_{1}$. If the direction of $\overrightarrow{Q}$ is reversed,the resultant has a magnitude $R_{2}$. The value of $(R_{1}^{2} + R_{2}^{2})$ is:

The resultant of two vectors $\vec{P}$ and $\vec{Q}$ is $\vec{R}$. If $\vec{Q}$ is doubled,the new resultant vector is perpendicular to $\vec{P}$. What is the magnitude of $\vec{R}$?

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.

The angle made by the resultant vector of two vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}-7 \hat{j}-4 \hat{k}$ with the $x$-axis is (in $^{\circ}$)

For two vectors $\vec{a}$ and $\vec{b}$,if $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$,then the angle between $\vec{a}$ and $\vec{b}$ is:

The sum of the magnitudes of two forces acting at a point is $16 \,N$. If their resultant is normal to the smaller force and has a magnitude of $8 \,N$, then the forces are: