If three vectors have equal magnitude i.e. $A=B=C$, then the angle between $\vec{A}$ and…
If three vectors have equal magnitude i.e. $A=B=C$, then the angle between $\vec{A}$ and
$\overrightarrow{\mathrm{C}}$ is $^{\prime} \alpha^{\prime}$. If $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}=0$, then the angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{C}}$ is ' $\beta^{\prime}$, then $\frac{\alpha}{\beta}$ is
$\frac{2}{3}$
$\frac{2}{1}$
$\frac{1}{2}$
$\frac{3}{2}$
Solution
The problem seems to involve vector algebra, where the magnitudes of three vectors are equal and there are questions regarding the angles between them.
Given that the vectors have equal magnitudes and their dot products might be involved, we can use the formula for the dot product between two vectors:
$\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}| \cos \theta$
where $\theta$ is the angle between vectors $\vec{A}$ and $\vec{B}$, and $|\vec{A}|=|\vec{B}|$ since the magnitudes are the same. Without specific values or a more detailed breakdown of the question, the answer to the angle-related query, based on standard results from such problems, is: