Statement I :
Two forces $(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})$ and $(\overrightarrow{\mathrm{P}}-\overrightarrow{\mathrm{Q}})$ where $\overrightarrow{\mathrm{P}} \perp \overrightarrow{\mathrm{Q}}$, when act at an angle $\theta_{1}$ to each other, the magnitude of their resultant is $\sqrt{3\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}$, when they act at an angle $\theta_{2}$, the magnitude of their resultant becomes $\sqrt{2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}$. 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 :-
Correct Option: 2,
$\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}$
$\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{P}}-\overrightarrow{\mathrm{Q}}$ $\overrightarrow{\mathrm{P}} \perp \overrightarrow{\mathrm{Q}}$
$|\overrightarrow{\mathrm{A}}|=|\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}$
$|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)(1+\cos \theta)}$
For $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}$
$\theta_{1}=60^{\circ}$
For $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}$
$\theta_{2}=90^{\circ}$
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