Why are the projections of an area element onto the coordinate planes the components of its normal vector?

Question

I understand that the components of the cross product $\bf{u}\times\bf{v}$ are the area of the parallelogram enclosed by $\bf{u}$ and $\bf{v}$, projected onto the respective coordinate planes (e.g. the z-component of the cross product is the parallelogram area projected onto the xy-plane, etc.).

Is there a geometric reasoning why the vector resulting from these components is actually normal to the area element? Of course this can easily be shown algebraically by taking dot products of $\bf{u}\times\bf{v}$ with $\bf{u}$ and $\bf{v}$ which are zero, but if possible I'd like some kind of intuitive explanation that is not based on equations.

Answer 1

This follows up on the idea floated by Chris Lewis. It uses physical intuition to explain the plausibility of the cross-product formula for the area-normal vector.

Given a right-angled set of coordinate axes, place a point on each axis and at the origin, and from these vertices make a tetrahedron that has three triangular faces on coordinate planes and one slanted transparent triangular face $S$ . Pick any unit vector $\vec n$ and imagine a uniformly intense beam of sunlight consisting of parallel rays travelling in the direction of $\vec n$.

1. The flux of solar energy entering through face $S$ is equal to the foreshortened area that $S$ presents when viewed by an observer looking in the direction $\vec n$. It also equals the sum of the exiting fluxes across each of the other faces, which also equals the sum of their foreshortened areas.

2. Each foreshortened area (on any face $F$) is $ n\cdot \vec A_F$ where $\vec A_F$ is the area-normal vector associated to that face; i.e. the vector whose magnitude is an area and whose direction points outward perpendicularly from that face.

3. The three coordinate faces have known area-normal vectors: $A_x \vec i, A_y\vec j, A_z\vec k$. (As you observed in your post, these three areas can each be computed using the 2x2 determinant formula.)

Thus (1) can be written as $$(4) \qquad \vec n \cdot \vec A_S=\vec n\cdot<A_x,A_y, A_z>$$

5. Since this identity is true for any choice of $\vec n$, we can now pick $\vec n$ in order to maximize the flux through $S$. It is geometrically clear that we want $\vec n$ to be the unit normal to the face $S$ to maximize the flux through $S$ (to minimize the effect of foreshortening).

On the other hand, the maximum value of the dot product $\vec n\cdot<A_x,A_y, A_z>$ occurs when $\vec n $ is parallel to $<A_x,A_y, A_z>$ ( a basic fact about dot products).

Thus the optimal chooice of $\vec n$, namely the unit normal vector to $S$, has the same direction as the vector whose components are computed by 2x2 determinants: the vector $<A_x,A_y, A_z>$.

Answer 2

It is explained exactly in this letter from Hamilton to Graves. He starts with the notion that there should be an axis $j$ perpendicular to the real axis and the imaginary axis of the complex numbers.

He then considers the product of two triplets $$(a_1+b_1i+c_1j)(a_2+b_2i+c_2j)$$ and is bothered by the $ij$ term. He ultimately develops the quaternion product which includes both the dot product and the cross product. The fact that $i$, $j$ and $k$ are perpendicular is required from the outset.

The algebraic skills of William Hamilton and the language of the day make it a challenging letter to read but, in the context of the resulting quaternions, the products of pairs of what we would call basis vectors are considered to be perpendicular to the pairs.