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Division of vector

  1. Apr 6, 2008 #1
    How can vector division be performed?
    Tell me the procedure how I would perform following expression:

    If A= 4i + 5j + 6k and B= 1i + 2j + 3k what will be A/B?
  2. jcsd
  3. Apr 6, 2008 #2
    Vector division is not well defined
  4. Apr 6, 2008 #3
    why is it not well defined? i want to know. explain it in detail plz..
  5. Apr 6, 2008 #4


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    First you are going to have to tell us what you think a "vector" is! There are several different (but equivalent) definitions. In fact there are several different definitions of "multiplication" for vectors. And only scalar multiplication includes a definition of division because there is no definition of multiplicative inverse for vectors.
  6. Apr 8, 2008 #5
    think about the following ...

    what is dividing ??

    supouse u have a set A, its elementes are a,b,c,...

    then supouse that u have an operation over A. say:

    c = a . b = b . a (for simplicity lets talk about commutative multiplication)

    where c in A, and a,b are any elements in A.

    the supuse that there is an unit element in A, say that this unit element its 1. this unit element
    is such a one that:

    a . 1 = 1 . a = a

    for any a in A.

    then, u can define what division is... if for any element a in A there is an element b such that:

    a . b = b . a = 1

    then, u can call this element b the inverse element of a, that is b=a^-1.

    now, the only remaining thing, its just to think of A as a set of vectors. and to thin of the multiplication

    a . b = c

    as an operation that leads two pairs of vectors to a new vector.

    the u must find the unit vector 1, and then u must prove that for any a in A there exist a^-1.

    the plus operation over vectors, has its properties. say



    a+(-a) = (-a)+a = 0

    so, in some way its an multiplicative structure over vector sets. but now there is the remaining
    property that finally defines multiplication.

    suppose, that in A there is a plus + operation. then a multiplication operation must follow the distributive

    ( a + b ) . c = a . c + b . c

    and that is... if u find a set of vectors A, with a plus operation a+b, and another operation a.b , such that holds:

    commutative summation

    exist 0 in A

    exist (-a) for any a in A
    a+(-a) = 0

    commutative multiplication
    a.b = b.a

    exist 1 in A
    a.1 = a

    exist (a^-1) in A for any a
    a.(a^-1) = 1

    (a+b).c=a.c + b.c

    then u have u multiplication operation over vectors

    best regards
  7. Apr 8, 2008 #6
    Yes, you can define vector multiplication and division that way (i.e., component-wise), but it turns out not to be nearly as interesting as the other ways of doing it (dot product and cross product) or matrix-vector multiplication. For example, what is the geometric interpretation of a component-wise product of two vectors?
  8. Apr 8, 2008 #7
    what is a component wise product ?? :)

    in fact I don´t know if a vector product exists.
    but, if it is so, I just gave the properties that it must carry out. I´m not an expert on the field, but those properties are the one that I saw on my algebra courses.

    best regards
  9. Apr 8, 2008 #8


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    Science Advisor

    No, you did not give "the properties it must have". There are two types of product that are defined for all (finite dimensional) vectors: scalar product: av where a is a scalar (a number) and v is a vector: you can think of it as "stretching" the length of the vector by a (if a>1) or "shrinking" v (if a< 1). Multiplicative inverses or division by a vector for such a product is not defined and could only be defined in the very restricted sense of requiring that u and v, in u/v, be in the "same direction" There are also "inner product" (basically the same as "dot product" of two vectors which does not result in a vector. Since that does NOT give a vector result, it cannot be divided by a vector.

    For 3 dimensions, you can define the "cross product" of two vectors. I think you could define "multiplicative inverses" or "division" for at least some vectors there, but, again, that is restricted to 3 dimensions. I was surprised at your sentence but then realized that your "algebra" course is probably not the algebra course I would think of in connection with vectors.
    Last edited by a moderator: Nov 8, 2008
  10. Apr 8, 2008 #9
    It's where you get the product vector by simply multiplying each component of the multiplicand vectors.

    The component-wise product fulfills all those properties.
  11. Apr 8, 2008 #10
    geometric algebra has a well defined, and very useful, vector multiplication and divison operation (geometric product, or clifford product). It's not closed, nor generally commutative, and produces a scalar and bivector (wedge) product. An example application of division is orthogonal decomposition with respect to a reference vector:

    x = \frac{1}{a} a x = \frac{a}{a^2} a x = \frac{a}{a^2} (a \cdot x + a \wedge x) = \hat{a} (\hat{a} \cdot x) + \hat{a} (\hat{a} \wedge x)

    This first part is the familiar vector projection operation, and the second is the orthogonal component.

    The basic rules for the multiplication(/division), assuming euclianian metric are:

    - associativity
    - linearity
    - [tex]a^2 = {\lvert a \rvert}^2[/tex]

    As a subject this ties together quite an astounding number of seemingly unrelated bits and pieces (complex numbers and quaternions, dot & cross product, linear equation solution, and much more.)
  12. Apr 9, 2008 #11
    I think the simple quesion is: "We are considering vectors in R^3 (vector vectors, not vectors as in 'vector spaces')- why is vector division not defined"?

    Since we have two types of vector multiplication- dot and cross product- we expect two types of division, one for each. We expect for three vectors a, b and c:
    And so vector division would allow us to say:

    For the dot product, c is a scalar. Given c and b for the dot product, can we determine a?
    Similarly for the cross product, given the product c and given b, can we find a?

    Or am I totally off-track?
  13. Apr 9, 2008 #12
    The geometric product does give a well defined vector divison for R^3. The cross product you mention is a special case (and makes a lot more sense in the context of a complete vector product).
  14. Apr 9, 2008 #13
    I won't pretend to understand what you have just said. I am assuming, incorrectly probably, that mdnazmulh is a first year student (or high school) and doesn't know about the complete vector product, neither do I, and only dot and cross product of vectors in R^3. That is the sense I got from the question. Why then is vector division not defined?
  15. Apr 9, 2008 #14
    Careful now. GA works in the Grassman-complete space --- remember that ab isn't a vector in R^3 if a and b are.
  16. Apr 9, 2008 #15
    I still think everyone is going too deep into the question. Look at the question:

    If A= 4i + 5j + 6k and B= 1i + 2j + 3k what will be A/B?

    That is:

    A=<4,5,6>, B=<1,2,3>, what is A/B? Like I said, I think he means vectors in R^3, and we are only considering dot and cross products- why is division not defined.

    Think back to when you were in first year, or high school, whenever you first came across vectors, three components, the unit vector along the x, y and z axes, i,j and k. You were taught vector algebra, that is, addition, subtraction, dot and cross product, but not division. I was told, division is meaningless for vectors. We could define a division, but it wouldn't be any use, so we don't. I am not a mathematician, only a second year student, but I am trying to get his question answered, which I would also like to know, but I think people are going too deep into the question.
  17. Apr 9, 2008 #16
    Partly because the question isn't that trivial. There is nothing to stop you from trying to define a multiplication and inverse --- i.e. a division. However, attempts to do so usually mess up some convenient feature that we would like, such as remaining in R^3 (GA) or having a basis independent result (component-wise multiplication). As far as things have uses go, that's never a reason to not do some maths :wink:
  18. Apr 9, 2008 #17
    Lol, I agree, but when we asked our prof in first year why there was no vector division, he said that it could be defined, but it isn't much use. I guess he was trying to shield us from complicated higher mathematics you guys are talking about.
  19. Apr 9, 2008 #18
    I'd guess that this prof was probably talking about the moore penrose "pseudoinverse" for matrixes. That is also well defined for vectors, and produces the scalar value one when multiplied by the original vector. It happens that this inverse is the same as the GA inverse, but without the rest of the context from that mathematics there isn't much to do with it.

    It's ironic that you are objecting to the "higher" math here. When I was back in high school I objected to the arbitrariness of the cross product (it doesn't even work for R^2) and seems very much like something that some mathematician pulled out of a magic hat. It's my expectation that within twenty years much of the basic vector algebra will be taught from a GA point of view ... it makes so much more sense that way. The texts that teach it now (like Hestenes's "New Foundations for classical mechanics") aren't exactly light reads, so some work is required to dumb it down. It's not a matter of teaching more advanced math, just doing things differently:


    - don't teach determinants, cramer's rule and matrix inversion as a starting point, teach the wedge product. The rest follows from that and is much easier to understand.
    - don't teach the cross product and dot products as special multiplication operations. Teach the GA product that incorporates these both, and allows consistent operation on higher dimensional object like planes, not just vectors, and not just R^3 (you'll need R^4 to deal with maxwell's equations once you take E&M but won't have the tools to do so).
    - introduce complex numbers as a special case of the Geometric product.
    - don't teach multivariable calculus for R^3 using only div and grad and then have to relearn it all to add one dimension (or work in R^2). Previously you'd have to use exterior calculus to get the general results and this is again an area that this simplifies.
    - ...

    It would be too easy to go on preaching on the subject (it's kind of exciting to find something that simplifies so much and then find that the subject has been around untaught (even to your prof probably)), so I'll stop. I'm also not going to answer the specific question of "what is" the A/B of your question since the context to answer that requires some work.

    Instead here are some pointers to actually learning the subject, what it is, and how to work with it:

    http://www.science.uva.nl/ga/tutorials/ [Broken]

    It's a interactive GA tutorial/presentation for a game programmers conference that provides a really good intro and has a lot of examples that I found helpful to get an intuitive feel for all the various product operations and object types.

    Even if you weren't trying to learn GA, if you have done any traditonal vector algebra/calculus, IMO its worthwhile to download this just to just to see the animation of how the old cross product varies with changes to the vectors.

    You have to download the GAViewer program (graphical vector calculator) to run the presentation. Once you do that you can use it for other calculation examples. See:

    http://www.geometricalgebra.net/downloads/gaviewerexercises.pdf [Broken]

    for some examples of how to use this as a standalone tool (note that the book the drills are from use a different notation for dot product (with a slightly different meaning and uses an oriented L symbol dependent on the grades of the blades).

    2) One possible starting place is lect1.pdf from the following:


    3) This one is pretty readable too:


    4) http://en.wikipedia.org/wiki/Geometric_algebra

    (I've dumped some content in there as I tried to learn it .. needs a lot of work)

    with regards to myself. I'm not a higher mathematician ... just a dumb computer programmer whos been through engineering school. I've stumbled on the subject after pulling out my old E&M books (compensating for boredom at work) and give them a re-read to understand them in the way I wished I'd had time to so back in school (like the high school days where all of physics was obvious and one didn't have to memorize any formulas because you could figure it out from first principles).
    Last edited by a moderator: May 3, 2017
  20. Apr 9, 2008 #19
    Whilst I agree that GA is quite a good pedagogical road, and certainly does contain some pretty neat formalism, the question of whether it actually simplifies anything is much more contentious. My perspective is from working with the people in the Cambridge group on applying GA --- so it's not entirely vacuous mouth/finger movements. My feeling is that it is an excellent advance for engineers, and people who use vectors and algebra and stuff to calculate. But less use for physicists and mathematicians, who are usually quite a lot more exacting in their needs, and are looking quite hard at foundations. In particular, quite a lot of the content in GA are actually well known in maths, under different names and not as unified --- but that's because people haven't found it theoretically useful to unify them. Note that theoretically useful has almost no bearing on practically useful. So things like differential geometry, which is such a huge part of modern theoretical physics, is almost completely incompatible with GA. After working with it for so long, it seems that the unification achieved by GA is too superficial, and doesn't contain enough depth to really advance mathematics itself.
  21. Apr 9, 2008 #20
    I suppose that time will tell. From my point of view I found the subject of differential geometry unlearnable until I looked at it from a GA point of view. Doesn't look to me that it is incompatable either -- notatationally different, but describes the same sort of operations.
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