# Dot product cross product, where did they come from?

1. Jan 16, 2007

### dionysian

While taking linear algebra I never really understood where the dot and cross product came from. It seemed that they were just randomly defined. I know that the dot product is an inner product but why is it defined as: $$\vec v \cdot \vec u = v_1 u_2 + v_2 u_2$$? why not $$\vec v \cdot \vec u = 3v_1 u_2 + 2v_1 u_2$$? I do see how defining the dot product as $$\vec v \cdot \vec u = v_1 u_2 + v_2 u_2$$ helps solve problems in physics, but is that the only reason why the dot product is defined that way?

I have pretty much the same question about the cross product, i see how the definition of the cross product helps solve physics problems, but it still seems just made up to me. Is this the case or is there some other deeper mathematical meaning behind the cross and dot product that eludes me?

Also why are they called products? What makes these two operations multiplication?

2. Jan 16, 2007

### HallsofIvy

Staff Emeritus
I'm not sure what you mean by "randomly" defined. The crucial point about the dot product is that it gives a way of finding the angle between two vectors and, especially, the fact that the dot product of two vectors is 0 if and only if the two vectors are perpendicular.

Of course, the cross product is defined so that the cross product of two vectors is perpendicular to both. In addition, if u, v, and w are vectors representing the 3 concurrent sides of a parallelopiped, then the volume is given by u.(vxw).

3. Jan 16, 2007

### slider142

I recommend reading Spivak's "Calculus on Manifolds" for illumination on the definitions of these two products.

4. Jan 16, 2007

### mjsd

A dot product is a specific case of a thing we normally refer to as inner product. It is a specific case because the associated metric is simply the identity (ie. the basis vectors are (1,0) and (0,1)) In a generalised case, the inner product case, you need not use these as you basis vectors and indeed your "dot product" should really look like
$$\vec u \cdot \vec v = \sum_{ij}^n g^{ij} u_i v_j$$
where $$g^{ij}$$ is your metric which can be any $$n\times n$$ matrix.

now, why dot product/inner product? Ok we have a vector space, we can have linear operators, operating on this space... etc... but we do not really have a mean of defining "geometry" is this space yet. An inner product is a tool to help one introduces the concept of lengths and angles.... *especially* in vector spaces of higher than 3 dimensions! Also, if you look at it carefully you realise that an inner product always produce a complex number as a result. Therefore it is a map $$inner: V\times V \rightarrow \mathbb{C}$$ . Now, in order to make this mapping to of some use to us...like the "geometry" thing I mentioned above, it turns out that the formula you are given suits the purpose.

By introducing the inner product to our vector space, we have now what we called an inner product space, now the inner product that we've introduced must satisfy the inner product axioms for it to be a proper inner product (or dot product in that special case). Ok, once you have got this you can now start talking about geometry. Why? well, think about the 2D case, $$x^2+y^2=L^2$$ where $$L$$ here is the length of vector (assuming a standard metric, ie. dot product, metric is identity), It would be nice if we can do similar thing in 4D, 5D... etc. eh? Although we may not call the resultant complex number from the dot product a "length" as such, it turns out that at least things are consistent (geometry wise). For instance, you can proof (using Cauchy-Schwarz ineq) that the "angle" between two vectors (in 4D, 5D.... ) will always obey $$|cos \theta| \leq 1$$.

As far as cross product goes, note that it is only defined in 3-dimension. This, I believe, has an origin from physics, eg. right hand rule...etc. Not surprising that it helps solve physics problem!!

5. Jan 17, 2007

### Swapnil

I thought that the cross product had a pretty good mathematical basis. Isn't it a special case of the wedge product?

6. Jan 17, 2007

### Swapnil

As a student of physics, I would say that this is a good enough reason. By defining the dot product in this way, we get an equivalent form:

$$\vec{a}\cdot\vec{b} = ||a|| ||b||\cos{\theta}$$,

which is extremely useful for physicists. Usually, but not exclusively, physicists use the dot product in the following way:

$$\text{comp}_{\vec{B}} \vec{A} \equiv \vec{A}\cdot\hat{B}$$

where $$\text{comp}_{\vec{B}} \vec{A}$$ is called component of $$\vec{A}$$ in the direction of $$\vec{B}$$. In this way, phycisists can know how much "influence" that vector $$\vec{A}$$ has in a given direction $$\vec{B}$$

As far as dot products go, it makes sense to call a dot product a "product" because it becomes your good old familiar product in 1 dimention. i.e. if your vectors are just 1 dimentional. For example, if $$\vec{u} = <u_1>$$ and $$\vec{v} = <v_1>$$ then $$\vec{u}\cdot\vec{v} = u_1v_1$$

Last edited: Jan 17, 2007
7. Jan 17, 2007

### neurocomp2003

dot/cross come from geometry.

8. Jan 17, 2007

### mjsd

In some situations, a wedge product can look like a cross-product very much, but I wouldn't go as far as saying that it is a sub-set of the generalised wedge product. Notably, we have for wedge products
$$(a \wedge b) \wedge c = a \wedge (b \wedge c)$$
while the same is not true for cross-product in general
$$(a \times b) \times c \neq a \times (b \times c)$$
i.e not associative

As far as vector properties are concerned, another thing to note is that the resultant vector you get from a cross-product is not a proper/polar vector (those you usually work with), it is a pseudovector (or an axial vector). A pseudovector remains invariant under improper rotation (eg. an inversion) For instance, for proper vector $$\vec u \rightarrow -\vec u$$ and $$\vec v \rightarrow -\vec v$$ under inversion, but for an axial vector defined by $$\vec a = \vec u \times \vec v$$, then under inversion it goes to itself
$$\vec a = \vec u \times \vec v = -\vec u \times -\vec v.$$

I believe that when Gibbs (i think is Gibbs, but I could be wrong) invented the cross-product he probably has the right hand rule in mind, but of course I can't be sure of that for I have never met Gibbs in person

I for one would want to know more about the true origin of the cross product too.

Last edited: Jan 17, 2007
9. Jan 18, 2007

### HallsofIvy

Staff Emeritus
Another way of generalizing the cross product (which really reduces to the wedge product): Define the "alternating tensor" in n dimensions, eijk...p to be 1 if ijk...p is an even permutation of 123...n, -1 if ijk...p is an odd permutation of 123...n, 0 if ijk...p is not a permutation of 123...n. The "cross product" of n-1 vectors is given by vi= eijk...pvjvk...vp. In 3 dimensions that is the usual cross product.

10. Jan 18, 2007

### Staff: Mentor

The development of the cross product is attributed to Josiah Gibbs and Oliver Heaviside, both of whom worked independently.

http://www.rtis.com/nat/user/jfullerton/school/math251/cproduct.htm

http://en.wikipedia.org/wiki/Josiah_Gibbs#Later_years
http://www.cgafaq.info/wiki/Cross_Product [Broken]

For those interested in Grassman Algebra, this might be of interest.
http://www.ses.swin.edu.au/homes/browne/grassmannalgebra/book/#Book

I think we should encourage Browne to finish his book.

Last edited by a moderator: May 2, 2017
11. Jan 18, 2007

### matt grime

You're looking at it from the wrong direction. This is no surprise, and is a by product of how mathematics is taught (i.e. with no emphasis on how things were discovered).

Remember, that in mathematics, things are presented backwards. That is definitions are almost always presented before the results we can deduce from the definitions. Those results really are what implied those were the correct definitions we wanted to choose.

So the reason for choosing that definition is precisely because it is useful.

A product (multiplication) is an operation from AxA to B, taking (x,y) to the 'product' of x with y. That's all.

12. Jan 18, 2007

### mr_homm

Responding mostly to mjsd:

Pseudovectors are also sometimes called bivectors in the 3d case (in case someone is googling for further information). The main difference between wedge and cross is that wedge products produce multicomponent objects (essentially antisymmetric tensors). It is interesting to consider HOW MANY distinct multicomponent objects can be created in this way. For n-dimensional space, the number of k-component basis vectors is Binom(n,k). (This is because the antisymmetry of the wedge product kills off anything with a repeated basis vector, so you must choose k distinct basis vectors from among n without replacement, which of course leads to the binomial coefficient). This means that in general the dimensions of the spaces of k-vectors and (n-k)-vectors are equal by the symmetry of the binomial coefficients. In particular, the dimension of the space of (n-1)-vectors is the same as the dimension of the original space of vectors (a.k.a. 1-vectors).

This means that one could set up a linear mapping of (n-1)-vectors onto the space of 1-vectors. In a general dimension n, an (n-1)-vector is of course the wedge product of n-1 vectors, so one would need to first wedge together n-1 vectors and then map back to the original vector space to get some kind of analog of the cross product.

On the other hand, to get something that resembles the simple product we are all familar with since grade school, you must multiply TWO vectors, hence the answer will be a 2-vector. Dimension n=3 is special in that the 2-vectors (a.k.a bivectors or pseudovectors) are the (n-1)-vectors, so you can define the cross product of two vectors by taking their wedge product and then mapping back from the bivectors to the ordinary vectors. It is obvious that this will only work in 3 dimensions.

It is precisely this extra mapping step that messes up the inherent associativity of the wedge product. A^B^C would be a 3-vector, but if you insert a mapping M back to the vectors at each stage, M(M(A^B)^C) gives the cross product. It is now absolutely obvious why you should not expect the cross product to be associative, since regrouping the product would move some vectors in or out of the argument list of the M function.

@ matt grime:

Strongly agree! Spivak's Calculus on Manifolds is a beautiful example of this, because he spends roughly 90 pages working out what the proper definitions should be, and then once they are in place lots of major theorems come out as immediate consequences. The nice thing about Spivak is that he actually stops and points this out, so that the reader is sure to get this insight, instead of just a lot of unmotivated derivation.

--Stuart Anderson

13. Jan 18, 2007

### mjsd

The pseudovectors business aside (that was brought up not as an argument but just some info for those who may be unaware of it).

True there is a mapping that you can do to get "some kind of analog" of the cross product. But my point (which I wasn't too clear on) was that how can one calls Thing-A a "subset" (perhaps using subset was the wrong word...but just forget about its technical meaning for the moment, for I can't think of a better word right now) of Thing-B when Thing-A does not even obey a fundamental property of Thing-B??? (in this case, associativity)

It is almost like the situation that, say, I can derive a "?-product" from the inner product using some mapping BUT this "?-product" does not obey one of the inner product space axioms. Ok should I still call the "?-product" an inner product like object?

mm... now I know why using the word "subset" may have caused some confusions....

Anyway, whether your answer to the above is Yes/No, I do not believe it is such an important matter...

14. Jan 18, 2007

### mathwonk

adding to matts comments, to me a product is frequently a pairing which is distributive over addition.

probably i say this because i work mostly with rings as opposed to groups.

15. Apr 2, 2007

### Swapnil

As far as cross products, the form in which you see them (i.e. as a determinant of a matrix) can be derived by just assuming that it follows these four axioms (notice that #4 is the only one that might seem "unnatural"):

1) $$\vec{A}\times\vec{B} = -\vec{B}\times\vec{A};$$ //anti-symmetry
2) $$\vec{A}\times(\vec{B}+\vec{C}) = \vec{A}\times\vec{B} + \vec{B}\times\vec{C};$$ //distributivity
3) $$a(\vec{A}\times\vec{B}) = (a\vec{A})\times\vec{B} = \vec{A}\times(a\vec{B});$$ //scalar multiplication
4) $$\vec{A}\times\vec{B} = \vec{C}$$ where $$\vec{C}$$ is perpendicular to both $$\vec{A}$$ & $$\vec{B}$$ and its direction is given by the right hand rule; //perpendicularity

Then using these a\timesioms we can derive these basic properties of the cross product:
1) $$\vec{A}\times\vec{A} = -\vec{A}\times\vec{A} => \vec{A}\times\vec{A} = 0;$$ //self-product

2) For $$\vec{A} = a_1\hat{e_1} + a_2\hat{e_2} + a_3\hat{e_3}$$ and $$\vec{B} = b_1\hat{e_1} + b_2\hat{e_2} + b_3\hat{e_3},$$ you have

$$\vec{A}\times \vec{B} = (a_1\hat{e_1} + a_2\hat{e_2} + a_3\hat{e_3})\times(b_1\hat{e_1} + b_2\hat{e_2} + b_3\hat{e_3})$$

$$= a_1\hat{e_1}\times(b_1\hat{e_1} + b_2\hat{e_2} + b_3\hat{e_3}) + a_2\hat{e_2}\times(b_1\hat{e_1} + b_2\hat{e_2} + b_3\hat{e_3}) + a_3\hat{e_3}\times(b_1\hat{e_1} + b_2\hat{e_2} + b_3\hat{e_3})$$

$$= a_1b_1(\hat{e_1}\times\hat{e_1}) + a_1b_2(\hat{e_1}\times\hat{e_2}) + a_1b_3(\hat{e_1}\times\hat{e_3}) + a_2b_1(\hat{e_2}\times\hat{e_1}) + a_2b_2(\hat{e_2}\times\hat{e_2}) + a_2b_3(\hat{e_2}\times\hat{e_3}) + a_3b_1(\hat{e_3}\times\hat{e_1}) + a_3b_2(\hat{e_3}\times\hat{e_2}) + a_3b_3(\hat{e_3}\times\hat{e_3})$$

$$= a_1b_2(\hat{e_1}\times\hat{e_2}) + a_1b_3(\hat{e_1}\times\hat{e_3}) + a_2b_1(\hat{e_2}\times\hat{e_1}) + a_2b_3(\hat{e_2}\times\hat{e_3}) + a_3b_1(\hat{e_3}\times\hat{e_1}) + a_3b_2(\hat{e_3}\times\hat{e_2})$$

$$= a_1b_2(\hat{e_1}\times\hat{e_2}) + a_1b_3(\hat{e_1}\times\hat{e_3}) - a_2b_1(\hat{e_1}\times\hat{e_2}) + a_2b_3(\hat{e_2}\times\hat{e_3}) - a_3b_1(\hat{e_1}\times\hat{e_3}) - a_3b_2(\hat{e_2}\times\hat{e_3})$$

$$= (a_1b_2 - a_2b_1)(\hat{e_1}\times\hat{e_2}) + (a_1b_3 - a_3b_1)(\hat{e_1}\times\hat{e_3}) + (a_2b_3 - a_3b_2)(\hat{e_2}\times\hat{e_3})$$

$$= (a_1b_2 - a_2b_1)\hat{e_3} - (a_1b_3 - a_3b_1)\hat{e_2} + (a_2b_3 - a_3b_2)\hat{e_1}$$

$$= (a_2b_3 - a_3b_2)\hat{e_1} - (a_1b_3 - a_3b_1)\hat{e_2} + (a_1b_2 - a_2b_1)\hat{e_3}$$

which is exactly the popular matrix form of the cross product that is given as its "definition" in most books.

Last edited: Apr 2, 2007
16. Apr 3, 2007

### leon1127

I thought it was because crossproduct was from the purely imaginary multiplication of quaternions which was to describe 3D space.

17. Apr 3, 2007

### Swapnil

I don't know much about quaternions so I can't really say anything. But I am sure there are many ways to make sense of the cross product. I just gave one way which can be used to do so.

18. May 14, 2007

### pkleinod

The following offers a slight change of perspective which I hope may be helpful.
One can understand the various products of two vectors in terms of a
geometric product that is associative and distributive over
addition but is not commutative wrt multiplication (ab does not necessarily equal ba).
In order to enable a geometric interpretation of the ensuing algebra, it is
taken as an axiom that the product of a vector by itself is a scalar.
Let a be the magnitude of the vector A. For Euclidean geometry, one usually takes
AA = A2 = a2. Now apply this axiom to the identity

(AB + BA) = (A+B)2 - a2 - b2.

Each term on the rhs is the square of a vector, hence the rhs is a scalar,
implying that (AB + BA) must also be a scalar. If we define

A.B := (AB + BA)/2
A/\B := (AB - BA)/2,

then the geometric product of two vectors, AB, can be written as the sum
of a scalar (the dot product, A.B) that is symmetrical in A and B and
a term A/\B (the exterior product) that is antisymmetrical in A and B
(A/\B = -B/\A):

AB = A.B + A/\B.

Let e1 and e2 be two unit vectors perpendicular
to each other and let U be a third unit vector lying in the plane defined
by e1 and e2. Vector U may be written as a sum of
components in the e1 and e2 directions:

U = cos(phi) e1 + sin(phi) e2.

The magnitude of the vector U is

U2 = 1 + sin(phi)cos(phi) (e1e2 + e2e1),

but, since U is a unit vector, the rhs must equal 1 for all values of phi.
This is only possible if e1.e2 = 0; i.e. the dot
product of perpendicular vectors is zero, which is just what we desire from
a geometric interpretation. This in turn implies that the
geometric product of perpendicular vectors is identical with the exterior
product:

e1e2 = e1/\e2,

and that the geometric product of perpendicular vectors is
antisymmetrical. Note that the square of the exterior product is -1:

(e1/\e2)2 = e1e2 e1e2
= -e1e2e2e1
= -1.

This shows that the exterior product of two vectors is neither a scalar
nor a vector. It is a bivector and can be interpreted as a segment
of directed area analogous to the interpretation of a vector as a
directed line segment.

If vector A is defined in terms of an orthonormal basis ei,
then A = (sum over i) aiei and
B = (sum over i) biei. Because the dot product
is distributive over addition (simple to prove!) and because
ei.ej = deltaij,

A.B = (sum over i) aibi.

So the definition of the dot product is not at all arbitrary.

Your second question concerned the cross-product. It is, in my opinion,
best to consider this not as a fundamental product but as an
ancillary recipe for obtaining the vector in 3D that is
associated with the exterior product. It is defined as the dual of
the exterior product:

A x B := -I (A/\B),

where I = e1e2e3 is called the pseudoscalar. If
you substitute the expressions for A and B in terms of their components
ai and bi into the rhs of this expression, you will get the usual expression for the cross product. End of story.

But perhaps there is a moral to the story: We should do ourselves a favour
by avoiding the cross product like the plague, except as a computational
tool. Physical quantities such as angular momentum should always be
defined in terms of the exterior product,
(a bivector or, if you insist, pseudovector). because a bivector
automatically has the correct symmetry properties, allowing one to
forget about axial and polar vectors. A vector is a vector is a vector and
does not come equipped with a label telling you where it came from! If
you walked into a room full of vectors, there would be no way of telling which
were axial and which were polar, but vectors could be distinguished from
bivectors immediately.

19. May 14, 2007

### mathwonk

the dot producthas as antecedent the formula for slope of perpendicular lines.

i.e. two lines are perpendicular if their slopes are negative recoprocals. if you rewrite this it says they are perpendicualr if the dot product of their direction vectors is zero.

it also comes from the law of cosines which generalizes the pythagorean theorem. i.e. pythagoras tells when two vectors are perpendicular, and the law of cosines gives the cosine of the angle between any two vectors in terms of their lengths and dot products.

cross products come from the determinant formula for the volume of the block formed by three vectors in 3 space.

the wonderful advance is that making these static formulas into dynamic operations gives them far more flexibility for computation.

20. Dec 28, 2007

### djc

physicists...

I've been scouring the web for a few hours trying to find out about pseudovectors and cross products, and this is the best I've found. Thanks for the good discussion.

I'll add a comment and a couple questions: My search began with trying to figure out what physicists mean when they say "a pseudovector is something like a vector, but it transforms in this different way..."
I think I've got it, so I'll summarize (recall that my goal was to understand what the physicists mean, not to find the most natural explanation):

Say V is an n-dimensional vector space, and let BASES be the set of ordered bases for V. Note that having a vector (by which I mean an element of V) is the same thing is having a map v from BASES to R^3 (the set of triples of real numbers) that satisfies a change-of-basis property: If M is the change-of-basis matrix from bases B1 to B2, then M v(B1) = v(B2).

Having such a map is equivalent to having a vector v, since a vector corresponds to the map which takes a basis to the coordinates for v in that basis. So we'll take this as the definition of vector, and define pseudovector in a parallel way:

A pseudovector is just like a vector, except that if B1 and B2 have opposite orientation (i.e., the change-of-basis matrix has negative determinant), insert a minus sign in the above equation). Note that for vectors and pseudovectors both, you determine a (pseudo)vector by specifying the coordinates in any one basis (i.e., specifying the value of the map at one point).

Cross products: Now suppose n=3, and V has an inner product. Let B be any orthonormal basis, and let u, v be vectors. Define u \cross v to be the pseudovector w defined by w(B) = u(B) \cross v(B). (In this equation, "cross" makes sense because we already have the cross product in R^3, which is due to the fact that R^3 has a canonical orientation.) It's not too hard to check that this makes w a well-defined (i.e., it doesn't depend on the original choice of orthonormal basis) pseudovector.

A final comment: A lot of this probably makes more sense in the context of differential geometry, where defining a (tangent) vector (at a point) to be a map from coordinate systems in a neighborhood of the point to R^3 which satisfies a change-of-coordinate rule is actually a reasonable way to define tangent vectors. Also, I guess if your manifold is orientable (does the universe have a natural orientation?) you could fix an orientation once and for all, and then end up with vectors rather than pseudovectors (if you want) in your cross-product.

My questions:
Did I miss anything?
Does anyone know of an exposition of this approach which won't make a mathematician-in-training cringe? (The problem with physicists is they're talking about such cool stuff, but understanding them is so hard.)
...especially an exposition that explains the equivalence of this approach with the exterior algebra approach? (I suppose this will be just like the proof that the mathematician's and physicist's definition of tangent vector are equivalent.)

-David