12 September 2007

geometric interpretation of vector products

Yesterday, a student asked me if there was a geometric interpretation of the dot product of vectors. I had been talking about the geometric interpretation of the cross product, which is usually given as follows: given two vectors a and b,

  • the magnitude of a × b is the area of the parallelogram determined by a and b, and

  • the direction of a × b is given by the right-hand rule
  • .

That's all well and good, although I was at one point asked what good the cross product was. This is a tricky question to answer, because it is not immediately obvious to students of calculus why one would want to find a vector that is orthogonal to two given vectors; furthermore, the definition of the cross product is a lot more complicated and mystifying than the definition of the dot product, and I suspect they want something more obviously useful in return for that.

Anyway, I couldn't think of a geometric interpretation of the dot product, in the sense that this student seemed to want. The Wikipedia article says the following: Since |a|cos(θ) is the scalar projection of a onto b, the dot product can be understood geometrically as the product of this projection with the length of b. That's true, and I did say that, but what good is it? The two vectors that we're saying ab is the product of the lengths of are in the same direction, so they don't naturally define an area. Is there a geometric interpretation of the dot product that doesn't feel so contrived? (This is sort of a vague question, as some people might not find what I just said contrived.)

We then moved on to the Scalar triple product a ⋅ (b × c), which created other frustration; it's weird to write this thing on the board and not mention that it's symmetric under even permutations of the arguments, because it's really the determinant of the 3-by-3 matrix containing those entries. I mumbled some words about how that sort of determinant comes up in changes of variables in multiple integrals, though; I was thinking of the Jacobian.

Of course, there's the classical claim that "you can only define a cross product in one, three, or seven dimensions", which I didn't mention, because nobody asked "can you define a vector product in two dimensions?" -- I would have mentioned it if they've asked.

But the cross product has that nice geometric interpretation. How do you continue with that? My officemate reminded me that the determinant of the "matrix"

\det \begin{pmatrix} \hat i & \hat \j & \hat k & \hat l \\a_1 & a_2 & a_3 & a_4 \\b_1 & b_2 & b_3 & b_4 \\c_1 & c_2 & c_3 & c_4 \end{matrix}

(where i, j, k, l are the unit vectors in four-space) is the 3-volume of the parallelepiped spanned by the three vectors (a1, a2, a3, a4), (b1, b2, b3, b4), (c1, c2, c3, c4). So it seems that we can define some sort of "cross product" in this sense in any number of dimensions; in Rn it'll take n-1 arguments. This is actually the wedge product in disguise.

But then what's the "cross product" (in this sense) of a single vector in 2-space? It's not the vector itself; if you write the "determinant" it's

\det \begin{pmatrix} \hat i & \hat \j \\a_1 & a_2 \end{pmatrix} = a_2 \hat i - a_1 \hat j

and so applying this operation to the vector (a1, a2) returns (a2, -a1). The operation takes in a vector and spits out its rotation by 90 degrees. This makes sense in retrospect; in the n-dimensional it spits out a vector which is orthogonal to the n-1 input vectors.

It just seems strange to be calling rotation by 90 degrees a "product"...


John Armstrong said...

The problem is that you're still trying to think in terms of "vectors". Okay, here's what the cross product really is:

A vector is a way of measuring an amount of an oriented 1-dimensional subspace of a vector space. That is, it picks out an oriented line through the origin and specifies a length.

A "bivector" is a way of measuring an amount of an oriented 2-dimensional subspace of a vector space. That is, it picks out an oriented plane through the origin and specifies an area.

Two (non-parallel) vectors give a bivector. Take the plane their lines span, orient it by the order the vectors come to you in, and take the area of the parallelogram spanned by the two lengths. Given the vectors (v,w) we call this the "wedge product v^w.

Similarly there are trivectors, and so on. In general, the space of k-vectors in an n-dimensional vector space is nCk (shouldn't be too hard to see)

So something funny happens in three dimensions. The space of vectors has dimension 3, and so does the space of bivectors! We can go back and forth between vectors and bivectors, picking one that's "perpendicular" to the other so that the trivector they make has volume 1 and is oriented the same way as the space. We call this procedure the "Hodge dual".

The cross product is the Hodge dual of the wedge product.

Chris said...

I think the difficulty arises from the order in which these subjects are tackled. Start talking about vectors and vector spaces before talking about coordinate systems. Talk about how to specify vectors and let this lead into the concepts of defining a basis vectors and how to define vectors in terms of a specific basis. The dot and cross products will then fall out as the natural ways to handle the definition of arbitrary vectors in terms of a basis.

Isabel said...

John, you make a good point; your words sound familiar (I learned this stuff at one point) but there's no way I would have thought of that on my own.

John Armstrong said...

I didn't have time to include this before, but that viewpoint also explains your conjecture about the "cross product" of a single vector in a 2-dimensional space. In such a space, the space of vectors is dual to itself. If you want to make a rectangle of area 1 with the right orientation, you rotate by 90 degrees and invert the length. Then the Hodge dual inverts the length again, giving the vector you saw.

Anyhow, don't worry about not having thought of it yourself. I'm (among other things) a geometer, and you're a probabilist/statistician/combinatoricist... I have no idea what your comment the other day about 2.11^n coming up in relation to trees is about, but you asked us not to ask you.

Aaron said...

Since |a|cos(θ) is the scalar projection of a onto b, the dot product can be understood geometrically as the product of this projection with the length of b. That's true, and I did say that, but what good is it?

Tremendous good! This kind of projection shows up in physics all the time. For example, consider an electric current flowing through a flat surface. The current is represented by a current density vector, whose magnitude is in units of current/area, and whose direction is the direction of the current. The surface is represented by an area vector, whose magnitude is the area of the surface, and whose direction is normal to the surface. To find out how much current is flowing through the surface, you project the current density vector onto the area vector---because you want to count the part of the current flowing through the surface and ignore the part flowing along it---and then multiply the projected vector's magnitude by the area of the surface, yielding a scalar with units of current/area * area = current.

Or consider a boxcar that's stuck on a track, so it can only move along the x axis. If you push on the boxcar in the (x+y) direction, you can see that half of your work will be wasted, because the boxcar can't move in the y direction. The dot product is how you'd translate this intuitive notion into formal math.

Scott Carter said...

I went through this the other day, too. My answer is more devious. The function xy is bilinear; that it is, is the distributive law. Area is also bilinear. I am talking about counting the number of squares in a rectangle to begin with. But area is also signed, we just are in the habit of ignoring the sign. GREEN SIDE UP!

Consequently, planar area is a bilinear alternating form. To keep track of the sign we introduce up and down. So now the plane lives in a space and divides it into up and down. Clearly we want ixj=k, etc so that the signs workout. So the cross product is the unique vector valued anti-symetric bilinear form, that takes specific values on i x j, j x k, and k x i.

Matt Noonan said...

I think that John's description of k-vectors (a really useful concept that I wish was more well-known) and Aaron's description of measurement really get at the heart of the matter here. For some reason, antisymmetric objects (k-vectors, k-forms, differentials, symplectic structures) seem to be more geometric than symmetric objects (dot product, Riemannian metrics, the shape operator), which appear to have more to do with measurement.

I think this is reflected in the fact that the cohomology of the tensor powers of T*M is the same as the de Rham cohomology, where we ignore symmetric powers. So at least the topology doesn't see symmetric things, in some sense. Similarly, every manifold carries a nonvanishing metric but not every manifold carries a nonvanishing 2-plane field, which is the antisymmetric version of a metric.

I wish I had more insight into this issue. I usually try to geometrize everything, but in a reversal of blogging roles, the dot product often turns into "correlation" in my head!

Anonymous said...

If a vector in R^{3} is viewed as an infinitesimal rotation, the cross-product can be interpreted in terms of the composition of infinitesimal rotations.

The triple product computes the oriented volume of a parallelopiped.

The ratio of the dot product of vectors a and b to the norm of b is the length of the shadow a makes on b.

TMS said...


It's really a very important question, i scratched my head and asked a lot of Phd's about it none of them give me the answer, after that and after a long nights fighting with Vector spaces and algebra I figured out it, shorty, Matrix is actually complex rotational & scaling & deformation of the vectors, and cross product as you said, is just a simplest one !

Anonymous said...

ab = a dot b + a wedge b
(a wedge b)starred = a cross b

Can be generalised.