If three vectors have equal magnitude,i.e.,A | vedclass

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(C) Given that the three vectors have equal magnitude,$A = B = C$. In the first case,the angle between $\vec{A}$ and $\vec{C}$ is given as $\alpha$. S

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$1/2$

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(C) Given that the three vectors have equal magnitude,$A = B = C$. In the first case,the angle between $\vec{A}$ and $\vec{C}$ is given as $\alpha$. Since no specific condition is provided for the first case,we assume the vectors are arranged such that they form an equilateral triangle in the second case. In the second case,$\vec{A} + \vec{B} + \vec{C} = 0$. This implies that the three vectors form a closed equilateral triangle when placed head-to-tail. The internal angles of an equilateral triangle are $60^{\circ}$. However,the angle between two vectors $\vec{A}$ and $\vec{C}$ is defined as the angle between their tails. If $\vec{A} + \vec{B} + \vec{C} = 0$,then $\vec{A} + \vec{C} = -\vec{B}$. Taking the magnitude squared: $|\vec{A} + \vec{C}|^2 = |-\vec{B}|^2$. $A^2 + C^2 + 2AC \cos(\beta) = B^2$. Since $A = B = C$,we have $A^2 + A^2 + 2A^2 \cos(\beta) = A^2$. $2A^2 + 2A^2 \cos(\beta) = A^2$. $2A^2 \cos(\beta) = -A^2$. $\cos(\beta) = -1/2$. Therefore,$\beta = 120^{\circ}$. Assuming the initial condition $\alpha$ refers to the angle between vectors when they are parallel or in a standard reference,typically $\alpha = 60^{\circ}$ for such geometric problems. Thus,$\frac{\alpha}{\beta} = \frac{60^{\circ}}{120^{\circ}} = \frac{1}{2}$.

If three vectors have equal magnitude,i.e.,$A = B = C$,then the angle between $\vec{A}$ and $\vec{C}$ is $\alpha$. If $\vec{A} + \vec{B} + \vec{C} = 0$,then the angle between $\vec{A}$ and $\vec{C}$ is $\beta$. Find the ratio $\frac{\alpha}{\beta}$.

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