Prove that in a $\Delta ABC$,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$,where $a, b, c$ represent the magnitudes of the sides opposite to vertices $A, B, C$,respectively.

(A) Let the three sides of the triangle $BC, CA$,and $AB$ be represented by vectors $\vec{a}, \vec{b}$,and $\vec{c}$,respectively (as shown in the figure).
We have $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
This implies $\vec{a} + \vec{b} = -\vec{c}$.
Taking the cross product with $\vec{a}$ on both sides: $\vec{a} \times (\vec{a} + \vec{b}) = \vec{a} \times (-\vec{c}) \Rightarrow \vec{a} \times \vec{a} + \vec{a} \times \vec{b} = -\vec{a} \times \vec{c} \Rightarrow \vec{0} + \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \Rightarrow \vec{a} \times \vec{b} = \vec{c} \times \vec{a}$.
Similarly,taking the cross product with $\vec{b}$ on both sides of $\vec{a} + \vec{b} = -\vec{c}$: $(\vec{a} + \vec{b}) \times \vec{b} = -\vec{c} \times \vec{b} \Rightarrow \vec{a} \times \vec{b} + \vec{b} \times \vec{b} = \vec{b} \times \vec{c} \Rightarrow \vec{a} \times \vec{b} = \vec{b} \times \vec{c}$.
Thus,$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$.
Taking the magnitude of these vectors: $|\vec{a} \times \vec{b}| = |\vec{b} \times \vec{c}| = |\vec{c} \times \vec{a}|$.
Using the definition of the cross product $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}| \sin \theta$,where $\theta$ is the angle between the vectors:
$|\vec{a}||\vec{b}| \sin(\pi - C) = |\vec{b}||\vec{c}| \sin(\pi - A) = |\vec{c}||\vec{a}| \sin(\pi - B)$.
Since $\sin(\pi - \theta) = \sin \theta$,we get:
$ab \sin C = bc \sin A = ca \sin B$.
Dividing the entire equation by $abc$,we get:
$\frac{ab \sin C}{abc} = \frac{bc \sin A}{abc} = \frac{ca \sin B}{abc} \Rightarrow \frac{\sin C}{c} = \frac{\sin A}{a} = \frac{\sin B}{b}$.
Hence,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.