[Math] Geometric proof of the Cross Product magnitude

Question

Most proofs of the magnitude of the cross product are algebraic in nature, I find I learn best visually / geometrically.

Is there a breakdown of the proof of the magnitude of the cross product using only geometry?

I don't have a (firm) background in linear algebra, so talk of the determinant or matrices isn't going to be terribly helpful for me. =/

I can establish some relationships with the triangles formed by the vectors and their cross product, but I don't want to butt my head against this problem, when I have so many chapters left to complete.

I'm working my way through Calculus 3 this summer to get ready for class in the fall.

Thanks.

Edit:
(My longer response to Brian's solution below)

Just to make sure I'm on the same page here:

• The notation for $a_i$ is the same as the notation my book uses for $a_1$, the portion of the $\vec{a}$ vector in the x direction.
• The "$\hat{i}$" is the unit vector for the x direction.
• Your "2D cross product" is the determinant of a 2×2 matrix (from the look of it); I was able to make that leap thanks to your definition in the comments.
• The "difference of two rectangles" would be the difference of the area of the rectangle formed by $a_i$*$b_j$ (the area of the first) and $a_j$*$b_i$ (second); The "$(a_i b_j – a_j b_i)$" portion.

Given all of that, your solution makes perfect sense. The leap from the book's definition and seeing the $a_i$*$b_j$ as the area of a rectangle is what really made the difference.

The only difference between this solution and the normal cross product, would be adding an extra dimension and the difference of two more rectangles.

Edit2:

And we can test for perpendicular vectors, because – when we construct our difference of rectangles – if the rectangles are equal (e.g. the vectors are perpendicular) then when we subtract the areas we get a sum of zero in each direction.

Answer 1

It will be easier to understand, perhaps, if we consider two vectors $\vec{a}$ and $\vec{b}$ restricted to the $x$-$y$ plane, so that $\vec{a} = a_i \hat{i} + a_j \hat{j}$, and $\vec{b} = b_i \hat{i} + b_j \hat{j}$. Then $\vec{a} \times \vec{b} = (a_i b_j - a_j b_i) \hat{k}$, and $\|\vec{a} \times \vec{b}\| = a_i b_j - a_j b_i$. A geometrical interpretation of this quantity can be seen as the difference between two rectangles:

(If you admit signed areas, it will still all work out.) We can then move two triangles over to the space left by the smaller rectangle:

And finally, we move two (differently shaped) triangles up to the previously uncovered part of the parallelogram:

Does that help?