If $\vec{a}+\vec{b}+\vec{c}=0,$ show that $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} .$ Interpret the result geometrically.

(N/A) Given,$\vec{a}+\vec{b}+\vec{c}=0$
$\Rightarrow \vec{b}=-\vec{c}-\vec{a}$
Now,$\vec{a} \times \vec{b}=\vec{a} \times(-\vec{c}-\vec{a})$
$=(\vec{a} \times(-\vec{c}))+(\vec{a} \times(-\vec{a}))$
$=-( \vec{a} \times \vec{c} ) - 0 = \vec{c} \times \vec{a} \ldots (i)$
Also,$\vec{b} \times \vec{c}=(-\vec{c}-\vec{a}) \times \vec{c}$
$=(-\vec{c} \times \vec{c})+(-\vec{a} \times \vec{c})$
$=0 - (\vec{a} \times \vec{c}) = \vec{c} \times \vec{a} \ldots (ii)$
From equations $(i)$ and $(ii)$,we get $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$.
Geometrical Interpretation:
If $\vec{a}, \vec{b}, \vec{c}$ are the sides of a triangle $ABC$ taken in order,then $\vec{a}+\vec{b}+\vec{c}=0$. The magnitude of the cross product of two vectors represents the area of the parallelogram formed by them. Since $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$,the areas of the parallelograms formed by any two of these vectors as adjacent sides are equal. This is consistent with the fact that these vectors form a triangle,and the cross products represent twice the area of the triangle formed by the vectors.