Last week I wrote this answer.

However, I don't feel it is complete. I make an arbitrary choice of unit vector perpendicular to u,v. But I can't see how to integrate this fact into the derivation.

I would like to complete the line of reasoning, as it motivates the crossproduct from the elementary task of constructing a vector perpendicular to two others.

Can anyone repair the narrative without escalating the complexity? Or get me to a point of understanding where I can make such a repair.

Here is a copy of the original (the original may change!):

There is indeed a geometric interpretation of $\bf{u}\times\bf{v}$ in terms of the areas of the projections of the parallelogram $\bf{P}$ spanned by $\bf{v}$ and $\bf{w}$ onto the coordinate planes.

I'll start from scratch.

Motivating problem: We wish to create a vector perpendicular to u,v, i.e. construct w s.t. $\bf{w}\cdot \bf{u} = \bf{w}\cdot \bf{v} = 0$.

Two equations in 3 unknowns: we can derive w as $\lambda(u_2 v_3 - u_3 v_2,$ [3&1], [1&2]$)$.

So let's define $\bf{u}\times \bf{v}$ as ($u_2 v_3 - u_3 v_2$, [3&1], [1&2]).

Observe $u_2 v_3 - u_3 v_2$ is just the signed area of the $(u_2, u_3)$, $(v_2,v_3)$ parallelogram. (2D determinant -- you can work it out with triangles). So we can rewrite our definition:

Define $\bf{u}\times \bf{v}$ as $(A_{yz}, A_{zx}, A_{xy})$ where $A_{\text{plane}}$ = signed area of u,v parallelogram projected onto that plane

Notice by symmetry, switching u and v just changes the sign of $u_p v_q - u_q v_p$, so we have:

$\bf{u}\times \bf{v} = -\bf{v}\times \bf{u}$

Notice also:

$\bf{i} \times \bf{j} = \bf{k}$

So we get our 'Right Hand Rule'.

Now it makes sense to ask: "Could we have skipped the algebra?" i.e. arrived at this definition purely from geometric insight. And the answer is yes!

Let n be the unit vector perpendicular to u,v using the Right Hand Rule to resolve ambiguity.

We can show that the ratio of the area of the uv-parallelogram to it's yz projection gives $\bf{n}_x$, similarly for xz$\to \bf{n}_y$ and xy$\to \bf{n}_z$:

Here is a simplification that will illustrate this...

Let's say we have y as depth, z going upwards. We have some near-flat plane (so the normal is pointing nearly straight up). Suppose we draw a unit grid on it. Now we are going to drop each point onto the xy-plane so $(x,y,z)\to(x,y,0)$ and calculate the change in area. Furthermore let's say we have rotated things so that our plane's normal vector has depth component = 0.

So alternatively we could set this up by imagining two xy-planes. And rotate one of them slightly around the y-axis, give it a unit grid, and "drop" this grid onto the other (axis-aligned) plane.

We want the area of a projected grid square.

It should be obvious that the depth-component (y) of any point on our plane is unchanged by this projection. And a simple calculation reveals that the x-component is just $cos(\theta)$, where $\theta$ is the angle between the normals, otherwise known as $\bf{k}\cdot \bf{n}=\bf{n}_z$, where $\bf{k}=(0,0,1)$.

So, the xy-area multiplier is $\bf{n}_z$ as required.

Let's write these parallelogram areas using $A$.

So we have: $A_{xy} = A \bf{n}_z$, sim. for y & z.

So $A\bf{n} = (A_{yz}, A_{zx}, A_{xy})$

(EDIT: Bold rendering incorrectly for $A_{xy}$ in the above line, anyone?)

So if we define $\bf{u}\times \bf{v}$ as $(A_{yz}, A_{zx}, A_{xy})$ then we have $\bf{u}\times \bf{v} = A \bf{n}$, i.e. $\bf{u}\times \bf{v}$ is perpendicular to u,v and of length $A$.


  • $\begingroup$ Which part are you not happy with? Choosing $\lambda=1$? The right hand rule? Also, is Oscar's answer to your other question on this topic not a good enough motivation for the cross product? $\endgroup$ – user137731 Jul 4 '16 at 2:16
  • $\begingroup$ @Bye_World, yes, essentially that is what I'm unhappy with. And both of those boil down to the same thing, because $\lambda=-1$ would give a left-hand rule. It's just that I have to make an arbitrary choice. Maybe this is connected with the alternating property. It just feels like the penny hasn't dropped somewhere. $\endgroup$ – P i Jul 4 '16 at 13:06
  • $\begingroup$ @Bye_World, Although his answer is insightful (and I am grateful for it), to me the motivation seems weak and contrived. i.e. We already have the cross product and we are trying to find a motivation. i.e. We start by rummaging around for an operator that meets certain criteria for no good reason! $\endgroup$ – P i Jul 4 '16 at 13:06

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