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I was reading a blog post on a simple derivation of the cross product. I learned how to determine the area of a parallelogram enclosed by two vectors $A$ and $B$.

First, here is the proof of the solution ($area = A_x B_y - B_x A_y$)

And here's the geometric implication of the solution.

It bewilders me that the geometric solution is simple but nonintuitive.

One commenter (Tim Poston) on the blog offered an explanation by using the following lemma, which he states can be easily shown using elementary plane geometry, without the use of right angles or base*height to find areas:

$C(u+w,v) = C(u,v) + C(w,v)$

where $u, v, w$ are arbitrary vectors, and $C$ is the area function of the parallelogram between 2 vectors.

Here's my illustration of an example implication of this lemma (visually simplified to triangles without loss of generality)

I was unable to prove this lemma, so I couldn't follow the argument. Any thoughts or direction with this lemma or the original problem would be appreciated.

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Warning: what you call "area" and "area function" refers to $\textit{oriented}$ area. Also, I hardly think this is "research level mathematics". –  Victor Protsak Jul 1 '10 at 5:34

2 Answers 2

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It is better to see this property if you stay with parallelograms and use the Cavalieri Principle. Here is page with an animation showing what I mean (clink on the link and accept).

There is also a demonstration of the first formula using triangles and again using only Cavalieri. There was an animation "proof without words" for it in the site "art of problem solving" but I can't find it anymore.

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Thanks, Cavalieri Principle was the key. I found the animation you were talking about: usamts.org/Images/Side9.swf –  Shaun Jul 3 '10 at 4:58

Draw a triangle with sides given by the vectors u, w, u+w. Draw three identical vectors v with their tails at the three corners of the triangle. Draw a new triangle with corners at the heads of the vectors. The total area of your figure can be seen as two parallelograms plus a triangle, but it can also be seen as a third parallelogram plus a congruent triangle.

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Thank you, it's much easier with parallelograms. –  Shaun Jul 3 '10 at 4:56

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