## An Example of a Parallelogram

Today I want to run through an example of how we use our new tools to read geometric information out of a parallelogram.

I’ll work within with an orthonormal basis and an identified origin to give us a system of coordinates. That is, given the point , we set up a vector pointing from to (which we can do in a Euclidean space). Then this vector has components in terms of the basis:

and we’ll write the point as .

So let’s pick four points: , , , and . These four point do, indeed, give the vertices of a parallelogram, since both displacements from to and from to are , and similarly the displacements from to and from to are both . Alternatively, all four points lie within the plane described by , and the region in this plane contained between the vertices consists of points so that

for some and both in the interval . So this is a parallelogram contained between and . Incidentally, note that the fact that all these points lie within a plane means that any displacement vector between two of them is in the kernel of some linear transformation. In this case, it’s the linear functional , and the vector is perpendicular to any displacement in this plane, which will come in handy later.

Now in a more familiar approach, we might say that the area of this parallelogram is its base times its height. Let’s work that out to check our answer against later. For the base, we take the length of one vector, say . We use the inner product to calculate its length as . For the height we can’t just take the length of the other vector. Some basic trigonometry shows that we need the length of the other vector (which is again ) times the sine of the angle between the two vectors. To calculate this angle we again use the inner product to find that its cosine is , and so its sine is . Multiplying these all together we find a height of , and thus an area of .

On the other hand, let’s use our new tools. We represent the parallelogram as the wedge — incidentally choosing an orientation of the parallelogram and the entire plane containing it — and calculate its length using the inner product on the exterior algebra:

Alternately, we could calculate it by expanding in terms of basic wedges. That is, we can write

This tells us that if we take our parallelogram and project it onto the – plane (which has an orthonormal basis ) we get an area of . Similarly, projecting our parallelogram onto the – plane (with orthonormal basis we get an area of . That is, the area is and the orientation of the projected parallelogram disagrees with that of the plane. Anyhow, now the squared area of the parallelogram is the sum of the squares of these projected areas: .

Notice, now, the similarity between this expression and the perpendicular vector we found before: . Each one is the sum of three terms with the same choices of signs. The terms themselves seem to have something to do with each other as well; the wedge describes an area in the – plane, while describes a length in the perpendicular -axis. Similarly, describes an area in the – plane, while describes a length in the perpendicular -axis. And, magically, the sum of these three perpendicular vectors to these three parallelograms gives the perpendicular vector to their sum!

There is, indeed, a linear correspondence between parallelograms and vectors that extends this idea, which we will explore tomorrow. The seemingly-odd choice of to correspond to , though, should be a tip-off that this correspondence is closely bound up with the notion of orientation.