The vector sum of two forces is perpendicular to their vector difference. In this case,the forces:

Two forces whose magnitudes are in the ratio $5:3$ are acting at a point at an angle $60^{\circ}$ simultaneously. If the resultant of the two forces is $35 \ N$,then the magnitudes of the two forces respectively are

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 $x$ and $y$ components of vector $\vec{P}$ have magnitudes $1$ and $3$,and the $x$ and $y$ components of the resultant of $\vec{P}$ and $\vec{Q}$ have magnitudes $5$ and $6$ respectively. What is the magnitude of $\vec{Q}$?

$A$ vector $\overrightarrow{A}$ when added to the sum of the vectors $(\hat{\imath}-2 \hat{\jmath}+2 \hat{k})$ and $(-2 \hat{\imath}+\hat{\jmath}-\hat{k})$ gives a unit vector along the $y$-axis. The magnitude of the vector $\overrightarrow{A}$ is

Among the given pairs of vectors,the resultant of two vectors can never be $3$ units. The vectors are