Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cross product intuition

  1. Jun 25, 2015 #1
    Hello, I hope this is the right forum section.

    I'm having trouble understanding how calculating the cross product arrives at the final result. When I do something simpler like multiplying a vector by a scalar, I can easily visualize in my head how each component "shrinks" or "grows".

    With the cross product, despite having seen animations of it and though I know how to apply it, it seems like witchcraft to me. No amount of articles or videos seem to help, they all miss the mark (or I the point).

    Do those of you familiar with the subject have an intuition of how the result vector is assembled during the calculation? Do the components of the two input vectors drag it back and forth until it stands upright?
     
  2. jcsd
  3. Jun 25, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I don't know what you mean by this last statement.

    By definition, the result of the cross product of two vectors is perpendicular to the plane formed by these two vectors. Changing the length of one or the other vector will not affect the plane formed by each; therefore, only the magnitude of the cross product will change.

    You can see this clearly by examining the cross products of the unit coordinate vectors:

    540b3860c7c84e4b3591b789ab0db4b8.png
    and

    e8355d9c4f09dcf706ebf8a54efa1dc1.png

    https://en.wikipedia.org/wiki/Cross_product
     
  4. Jun 25, 2015 #3

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yes, it is a bit witchcraft. In all honesty, there is nothing really to get about the vector product. There is no hidden intuition or anything. It is just a handy thing that we use often. If you have a plane in space, then there is a vector perpendicular to that plane, the cross product just denotes a way of picking out such a vector. There are various applications of the cross product, all of which use the concept in a different way. So as long as you understand the definition, it's ok. There is no intuition to get.
     
  5. Jun 26, 2015 #4
    Ok, thank you. Not being particularly talented at math, a response such as this from someone so experienced convinces me to lay the issue to rest; I'll just accept that it works and see it as a tool.
     
  6. Jun 26, 2015 #5

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The cross product is probably the simplest meaningful way to combine two vectors into one that's perpendicular to both. Of course, one can argue about the precise meaning of "simplest" and "meaningful" but still,...

    Consider 18 numbers ##A_{ij}## and ##S_{ij}## with ##i,j\in\{1,2,3\}##. Suppose that ##A_{ij}=-A_{ji}## and ##S_{ij}=S_{ji}## for all ##i,j\in\{1,2,3\}##. (We say that A is antisymmetric and S is symmetric). Now we have
    $$\sum_i\sum_j A_{ij}S_{ij}=\sum_i\sum_j (-A_{ji})S_{ji} =-\sum_i\sum_j A_{ji}S_{ji} =-\sum_j\sum_i A_{ij}S_{ij} =-\sum_i\sum_j A_{ij}S_{ij}.$$ In the third equality, I just used that i and j are dummy indices. This allows me to just swap them with each other. In the final step, I used that two summation sigmas can be interchanged. The above implies that ##\sum_i\sum_j A_{ij}S_{ij}=0##.

    Now let's say that we want to define ##x\times y## so that it's perpendicular to both ##x## and ##y##. Since two vectors u and v are perpendicular precisely when ##0=u\cdot v=\sum_i u_i v_i##, we will need the following results to hold $$\sum_i x_i(x\times y)_i=0 =\sum_i y_i (x\times y)_i.$$ The result ##\sum_i\sum_j A_{ij}S_{ij}=0## suggests a way to accomplish this. If we can find 27 numbers ##\varepsilon_{ijk}## with ##i,j,k\in\{1,2,3\}## such that ##\varepsilon_{ijk}## is antisymmetric under the exchange of any two indices, and define the cross product by
    $$(x\times y)_i =\sum_j\sum_k \epsilon_{ijk}x_j y_k,$$ then we can almost immediately see that ##\sum_i x_i(x\times y)_i## and ##\sum_i y_i(x\times y)_i## are both zero. For example,
    $$\sum_i x_i(x\times y)_i =\sum_i\sum_j\sum_k x_i\varepsilon_{ijk}x_j y_k=0$$ because ##x_ix_j## is symmetric with respect to the swap ##i\leftrightarrow j## while ##\varepsilon_{ijk}## is (for each ##k##) antisymmetric with respect to that same swap.

    Now how do we choose the 27 numbers ##\varepsilon_{ijk}##? It turns out that once we choose ##\varepsilon_{123}##, the other 26 numbers are fixed by the requirement that ##\varepsilon_{ijk}## is antisymmetric under the exchange of any two indices. The standard definition of the cross product is simply the choice ##\varepsilon_{123}=1##.
     
    Last edited: Jun 29, 2015
  7. Jun 28, 2015 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    If you are familiar with the application of computing the torque exerted by a force on a lever then the geometric definition of the cross product should seem natural - at least the magnitude A B sin(theta) should be obvious. Perhaps you are asking why the geometric definition agrees with the algebraic definition of the cross product, which computes it component by component. I don't know a simple way to see that visually. I think you have to understand it by seeing patterns in algebraic manipulations.
     
  8. Jun 28, 2015 #7
    Is this a way of saying that the cross product is a (2,1)-rank tensor over R3? (Sorry if mistaken, I'm a begginer with tensors)
     
  9. Jun 28, 2015 #8

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Define ##V=\mathbb R^3##. We can certainly use the cross product (which is a bilinear map from ##V\times V## into ##V##) to define a multilinear map ##T:V^*\times V\times V\to\mathbb R## by ##T(\omega,x,y)=\omega(x\times y)## for all ##\omega\in V^*## and all ##x,y\in V##. So the cross product indirectly defines a tensor. If you want to call it type (2,1) or type (1,2) appears to be a matter of convention. (I think Wald calls it (1,2) and Lee calls it ##{2\choose 1}##).
     
  10. Jun 28, 2015 #9

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    No. The cross product is instantly and always at right angles to both the original vectors.

    Forgive me if I am over-simplifying your question, but here is my two bits of "intuition":
    Suppose you wanted to know how fast something was flowing across a surface, from the inside of a volume to the outside. Consider a small rectangle or parallelogram part, R, of the surface whose sides at one corner are two vectors. The cross product of those two vectors tells you two things:
    1) What direction would be crossing the surface part R at right angles.
    2) How much surface area R represents.

    That is all you need to know about the surface there. You need to know the right-angle direction because flow parallel to R does not contribute to crossing the surface. Only the flow component at right angles to R counts. And of course, you need to know how much surface area R represents. The swapping the order of the vectors in the cross product points the cross product in the opposite direction. So if one order points toward the outside of the volume, the other order points toward the inside of the volume.
     
    Last edited: Jun 28, 2015
  11. Jul 13, 2015 #10

    chiro

    User Avatar
    Science Advisor

    It would serve you well for R^2 to look at the cross product in terms of multiplication of vectors as opposed to numbers.

    The roots of this sort of thing went back to Grassmann who was actually just a teacher who did this stuff alone (George Green was also a baker who did his theorem in his spare time as well).

    When we write a product c = a x b then we also (usually) want to have a way to divide something to remove a factor. So this means that c = (a x b) / b should equal a if everything is consistent and if it can be invertible (for example multiplying by zero can't be inverted).

    Grassmann was trying to do the same thing with vectors and was messing around with it quite a lot. Hamilton figured this out for a four dimensional structure called Quaternions but it was Grassmann and later people like Clifford and others that wanted to look a bit deeper.

    It turns out that you can actually do this with vectors (under certain conditions) and that there is a way of making sense of this sort of operation (multiplication if you like) with something that is known as a bi-vector which has a vector part and a scalar part.

    I'd suggest if you really want to learn the history and get rid of the "witchcraft" idea - you should go back to Grassmann and then look at how other mathematicians picked up where he left off and started to develop the geometric algebra (which is what it's called).
     
  12. Jul 19, 2015 #11
    I disagree. It is true that you can get away with just thinking of the cross product as something that multiplies the lengths of the vectors and gives you a perpendicular vector, according to the right hand rule, but that attitude comes from a fairly unquestioning point of view and is not appealing to those who don't like to just take someone else's word for when things should be done this way or that way. Also, there is a lot of pleasure to be had in figuring out the intuition behind things, regardless of whether you can save time by just accepting them. At some point, someone decided that the cross product was a good idea. The cross product wasn't just written in the sky, so someone just copied it down faithfully, and now we have to use it.

    So, I guess for now, I would have to second the chiro's recommendation to look into Grassman algebras.
     
  13. Jul 19, 2015 #12

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Regarding the reason that the cross product is always at right angles to the two original vectors. There might be a general method of constructing an orthogonal basis that applies. Or it might be the motivation of a general method. I would be surprised if it wasn't. I have seen it stated that the Gram-Schmidt orthogonalization equations gives the cross product equation. They would have to be multiplied to get the correct magnitude.
     
    Last edited: Jul 19, 2015
  14. Jul 24, 2015 #13
    Ok, here is my real answer, now that I am in a better frame of mind to discuss it. It really helps to have a lot of experience studying and using different related math concepts for the intuition to really sink in, but I think the general idea of it can be conveyed, although if I could draw pictures and put a lot of time into it, it would be clearer.

    The geometric definition is to take a perpendicular vector to the two given ones with magnitude equal to the area of the parallelogram they span.

    Often, the way this is computed is via the determinant formula. This formula does actually have some pretty good intuition behind it. In the interest of brevity, I'll probably still have to slide a bit under the rug here, so I may leave some of the mystery intact, but if you chase it down all the way, there's not much left.

    Let's say that Vol(u,v,w) represents the volume of the parallelopiped spanned by u, v, and w.

    So, with that in mind, let's say we have two fixed vectors, u and v, and we want to find a perpendicular to them. We can then define a function, f(w) = Vol(u,v,w). So, this function is telling us how big a parallelopiped we get when we take the parallelogram spanned by u and v and add w to turn it into a parallelopiped. So, then we can ask if w is sitting at the origin, where is the direction that this function increases the fastest?

    The answer is that perpendicular direction that we were looking for. The function doesn't even care if the tip of w moves parallel to u and v at all. In fact, f(w) is just the length of the projection of w onto the perpendicular vector to u and v times the parallelogram area of u and v. From what I have said, it should be plausible that f(w) = Aw⋅b. for some vector b, perpendicular to u and v and A is the parallelogram area.

    Now, we can just compute the volume Vol(u,v,i) = f(i) = i⋅bA to get the i-component for b. Similarly for the j-component and the k-component. That means if we can compute these volumes, we can find b.

    We can compute Vol(u,v,w) by using a determinant if we know about the geometric interpretation of determinants. If you work out the details, plugging in w = i, j,k, what you get here is just the standard determinant formula for the cross product.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cross product intuition
  1. Cross product (Replies: 15)

  2. Cross product! (Replies: 3)

Loading...