How can we resolve the area vector of a 2-D surface into its components?

Question

If we take a triangle marked by the coordinates (1,0,0), (0,1,0) and (0,0,1), as follows:

It appears that if we take the area of this surface to be a vector quantity, it is equal to the sum of three other area vectors which seem to have a vague relationship with the figure. Those three area vectors represent triangles whose vertices are the origin and the intercepts made by the figure with the axes.

Why is this true? Why does the vector sum of "shadows" make up the complete area?

Answer 1

When you say "take the area of this surface to be a vector quantity", I suppose you mean to represent the area with a vector $\mathbf v$ in the direction normal to the surface and with magnitude being the area?

Theorem:

If a triangle has vertices $A$, $B$, and $C$, then its area is given by $$\text{Area } = \frac 12 |\vec {AB} \times \vec{AC}|$$

Of course, the cross product of two vectors $\mathbf u$ and $\mathbf v$ returns a vector that is perpendicular to both $\mathbf u$ and $\mathbf v$. It follows that the "area as a vector quantity" that you want is precisely

$$\vec{Area} = \frac 12 \vec {AB} \times \vec{AC}$$

Now, going back to the question you were asking, if we let

$$\vec {OA} = \mathbf a \qquad \vec {OB} = \mathbf b \qquad \vec {OC} = \mathbf c$$

then the area of the slant surface $ABC$ as a vector is given by

\begin{align} \vec{Area}(ABC) & = \frac 12 \vec {AB} \times \vec{AC} \\ & = \frac 12 (\mathbf b - \mathbf a) \times (\mathbf c - \mathbf a) \\ & = \frac 12 \mathbf b \times \mathbf c - \frac 12 \mathbf b \times \mathbf a - \frac 12 \mathbf a \times \mathbf c \\ & = \frac 12 \mathbf b \times \mathbf c + \frac 12 \mathbf a \times \mathbf b + \frac 12 \mathbf c \times \mathbf a \\ & = \vec{Area}(OBC) + \vec{Area}(OAB) + \vec{Area}(OCA) \end{align}

which is what you wanted.