Two vectors $\vec A$ and $\vec B$ have magnitudes $2$ and $1$ respectively. If the angle between $\vec A$ and $\vec B$ is $60^o$, then which of the following vectors may be equal to $\frac{{\vec A}}{2} - \vec B$
Two vectors $\vec A$ and $\vec B$ have magnitudes $2$ and $1$ respectively. If the angle between $\vec A$ and $\vec B$ is $60^o$, then which of the following vectors may be equal to $\frac{{\vec A}}{2} - \vec B$
- A
- B
- C
- D
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Statement $I :$Two forces $(\overrightarrow{{P}}+\overrightarrow{{Q}})$ and $(\overrightarrow{{P}}-\overrightarrow{{Q}})$ where $\overrightarrow{{P}} \perp \overrightarrow{{Q}}$, when act at an angle $\theta_{1}$ to each other, the magnitude of their resultant is $\sqrt{3\left({P}^{2}+{Q}^{2}\right)}$, when they act at an angle $\theta_{2}$, the magnitude of their resultant becomes $\sqrt{2\left({P}^{2}+{Q}^{2}\right)}$. This is possible only when $\theta_{1}<\theta_{2}$.
Statement $II :$ In the situation given above. $\theta_{1}=60^{\circ} \text { and } \theta_{2}=90^{\circ}$ In the light of the above statements, choose the most appropriate answer from the options given below
Statement $I :$Two forces $(\overrightarrow{{P}}+\overrightarrow{{Q}})$ and $(\overrightarrow{{P}}-\overrightarrow{{Q}})$ where $\overrightarrow{{P}} \perp \overrightarrow{{Q}}$, when act at an angle $\theta_{1}$ to each other, the magnitude of their resultant is $\sqrt{3\left({P}^{2}+{Q}^{2}\right)}$, when they act at an angle $\theta_{2}$, the magnitude of their resultant becomes $\sqrt{2\left({P}^{2}+{Q}^{2}\right)}$. This is possible only when $\theta_{1}<\theta_{2}$.
Statement $II :$ In the situation given above. $\theta_{1}=60^{\circ} \text { and } \theta_{2}=90^{\circ}$ In the light of the above statements, choose the most appropriate answer from the options given below
- [JEE MAIN 2021]