Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces Also name the triangle formed by the forces as sides
$120^°$ equilateral triangle
$60^°$ equilateral triangle
$120^°, 30^°, 30^° $ an isosceles triangle
$120^°$ an obtuse angled triangle
The vectors $\vec{A}$ and $\vec{B}$ are such that
$|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$
The angle between the two vectors is
The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle between $A$ and $B$ is ....... $^o$
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