Why Vectors product the way it is?

[Amazement]

In regards to the products, most sources I read describing the basics of three dimensional vectors merely state what the definition of a cross product and dot product are without a geometric/trignometric proof.

I tried to prove the equation for the dot product (A.B=ABcosθ through the construction of a parallelogram with sides A and B with diagonal D.

Now to calculate D i arrive with an equation describing the diagonal as the addition of two cosines of two similar triangles formed by transposing A to the tip of B (i.e. half the parallelogram.)

D=Acosθ+Bcosγ​

Where θ is the angle of the right hand triangle with hypotenuse A and likewise for γ.

I personally cannot see from this point how you obtain the dot product equation, perhaps I am attempting it in completely the wrong manner. Thx in advance.

Oh and p.s. I am assuming its much more obvious in matrix form and I have read the matrix proof on PF but when I suggested matrices in class the professor was outraged that I made such a heretical suggestion.

 

[BruceW]

You can divide by a vector, but the result isn't another vector. In general a vector divided by a vector gives the sum of a scalar plus a bivector (either of which can be zero).

Dividing by vectors is actually useful. For two vectors a and b, the quotient (a/b) is an operator which performs rotations in the direction that takes a into b, by twice the angle between them.[*]

This comes from Geometric Algebra. A good reference is this book:
http://www.geometricalgebra.net/

[*] Well, up to a scaling factor, anyway. The point is that the quotient of vectors is a rotation operator.

Wow, I never knew this. Its above my level of mathematical knowledge, I guess :)

I tried to prove the equation for the dot product

For a parallelogram defined by vectors \vec{A} and \vec{B}, the squared length of the diagonal created is equal to (\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = D^2 (simple vector rules). So this will mean that A^2 + B^2 + 2\vec{A} \cdot \vec{B} = D^2

We can also work out the squared length of the diagonal geometrically (i.e. by just drawing the picture). You can do this by imagining another triangle fixed on the end of the parallelogram, then working out the lengths by using the angle. When you work it out, you get: A^2 +B^2 + 2ABcos(\theta) = D^2 And we can compare this with our other equation to get the result \vec{A} \cdot \vec{B} = ABcos(\theta)

Of course, this isn't a rigorous mathematical proof, but it is nice to see that it all works when you try it out yourself.

 

[AlexChandler]

In regards to the products, most sources I read describing the basics of three dimensional vectors merely state what the definition of a cross product and dot product are without a geometric/trignometric proof.

In math you do not need to prove a definition. We simply notice that there is this combination of symbols that we often see, so we decide to give it a name: "dot product" or "cross product". From these definitions you can prove that the cross product is perpendicular to the original vectors for instance, or that a dot product is zero only for perpendicular vectors.

 

[AEM]

I have come to this thread rather late, however I'd like to make a few comments. Vector analysis was worked out in the early 1880s by the American mathematical physicist, J. Willard Gibbs. In the period 1881-1884, he circulated a pamphlet he had written to interested people and in 1901 Yale University press published a book based upon lectures Gibbs gave at Yale University. As well as Gibbs, some aspects of what we now call vector analysis was also developed by Oliver Heaviside an others in England who were working over Maxwell's electrodynamics.

Looking at Gibbs' book "Vector Analysis", one can get an idea of what his thinking might have been in the early days when he was working things out. On page 62 it is pointed out that the vector product

\bf{A} \times \bf{B} = A B Sin{\theta}

represents the area of the parallelogram whose sides are A and B. After pointing this out he says,

"This geometric representation of \bf{A} \times \bf{B} is of such common occurrence that it might well be taken as the definition of the product."

He then goes on to state that "the vector product appears in mechanics in connection with couples." and later, "The product makes its appearance again in considering the velocities of the individual particles of a body which is rotating with an angular velocity given in magnitude and direction by A.

I would like to point out that in the nineteenth century, there were three other related algebraic systems being developed that became overshadowed by vector analysis. These were William Rowan Hamilton's Quaternions, Hermann Grassmann's algebra (now called Grassmann algebra), and William Kingdon Clifford's reworking of Grassmann's algebra into Geometric Algebra. All of these had their own version of the vector product (which I'll call the wedge product). It also was skew symmetric. The difference they had with Gibbs' vector product is that the order of multiplication serves to define an orientation of the plane determined by the two vectors, but not a third vector perpendicular to the plane. This means that the wedge product can be defined in a space of any dimension.

The improvement that Clifford made on Grassmann's algebra was among other things was to combine the scalar product and wedge product into one product that has two parts --one symmetric (the dot product) and one antisymmetric (the wedge part). This is analogous to combining real numbers and imaginary numbers into a single complex number.

In the 1960's, David Hestenes, now at Arizona State University rediscovered Clifford's Geometric Algebra and has written many articles and several books on Geometric Algebra. Many of these are available on the Internet at http://geocalc.clas.asu.edu/

Geometric Algebra is well worth your attention. It unifies a wide array of mathematics into one unified system including complex numbers, vector analysis, differential forms, and Pauli spinor algebra, just to mention a few. Geometric Algebra may be used in all branches of physics with some occasionally startling results. It currently is widely used in computer graphics, as indicated by the link given by Chogg.

Two more links where you can find additional information are

http://faculty.luther.edu/~macdonal/

and

http://www.mrao.cam.ac.uk/~clifford/

The reason that I've made this post so long is that I firmly believe that when it comes to physics, the more tools that you have to work with the better, and Geometric Algebra, while not usually taught in universities, is a powerful tool.

 

[nucl34rgg]

Hi,

I am wondering why the scalar product of two vectors the way it is, i mean why it wasn't ABtanθ instead of ABcosθ, or XAB, while X is any constant value, why mathematicians define it that way, the same story goes on for the vector product,
why we even had two types of products defined, while we don't have division defined,

and why it was "accidentally!" suitable as the best language to describe nature,
or i got the hole thing wrong and i have to think about it as it was defined as a new mathematics to best describe nature, in this case all of my questions will be answered as it had to be that way because we describe nature already and nature behave that way

my original thought that vector algebra is a topic of pure mathematics and it was developed not in mind describing nature, and after that physicists use it to describe nature

i need your opinion...

Conceptually, the dot product represents the multiplication of one vector by another projected onto it. What I mean by "projected' is qualitatively "how much along the other vector is this vector." This is the meaning of the inner product. It is a projection to test for how much "one thing" is "like" "another thing." Physically, with two vectors pointing in different directions, if you take the component of one of the vectors along the other and multiply them, you will have your dot product. Geometrically, this is taking the magnitude of one of the vectors and multiplying it by the cosine of the angle between them, since this value will give you the length of the component.

One cool thing you will see is that inner products aren't limited to what we normally call vectors (mathematicians call a lot of things vectors that you wouldn't normally think of as vectors). For example, the Fourier transform value at a particular frequency is an example of a function that gives u what is basically the inner product of your original function in the time domain with sine and cosine waves of the given frequency. In a sense, it shows you how much the frequency of the original signal is "like" that of the test frequency. (The peaks in the Fourier spectra are the frequency components of the source signal that made the most contribution.)

 

[bbbeard]

Hi,

I am wondering why the scalar product of two vectors the way it is, i mean why it wasn't ABtanθ instead of ABcosθ, or XAB, while X is any constant value, why mathematicians define it that way, the same story goes on for the vector product,
why we even had two types of products defined, while we don't have division defined,

and why it was "accidentally!" suitable as the best language to describe nature,
or i got the hole thing wrong and i have to think about it as it was defined as a new mathematics to best describe nature, in this case all of my questions will be answered as it had to be that way because we describe nature already and nature behave that way

my original thought that vector algebra is a topic of pure mathematics and it was developed not in mind describing nature, and after that physicists use it to describe nature

i need your opinion...

Well, here's my 2 cents. Maybe this is too technical, but the sooner you learn this, the better.

From the standpoint of physics, what makes something a vector is how it behaves under some set of coordinate transformations. In ordinary 3d space (as opposed to 4d Minkowski space, for example), the coordinate transformations we're usually interested in are rotations, which you are invited to think of in terms of rotation matrices. A vector is a "rank 1 tensor", i.e. a one-index object, say V_i where i=1,2,3 represent x,y,z. A rotation matrix is a two-index object, say R_{ij}. A rotation matrix acts on a vector, producing another vector

V'_i=\sum\limits_{j=1}^3 R_{ij}V_j

where the sum is over j=1,2,3. We have a convention, called the Einstein summation convention, which says if we write a product like R_{ij}V_j with a repeated index j, we are supposed to assume a summation over the repeated index, which allows us to write the expression above

V'_i=R_{ij}V_j

(That Einstein guy saved the world a lot of ink.) Anyway, we say any three quantities that transform under a rotation R_{ij} the way that the coordinates do is a "vector". Things like forces and momenta are vectors, but not every set of three quantities is a vector. For example, if we think about something like

X=(1/x,1/y,1/z),

this X is not a vector. That is, 1/x_i is not a vector -- it doesn't obey the vector transformation law V'_i=R_{ij}V_j.

We can also have two-index objects that obey a vector-like, or "tensor" transformation law. But we have to "rotate" both indices, that is, in order for T_{ij} to be a tensor, it has to obey a law like

T'_{ij}=R_{ik}R_{jm}T_{km}.

And we can continue making higher-order tensors, but each index gets its own copy of the rotation matrix. An important rank three tensor is the fully-antisymmetric tensor, the so-called "epsilon" tensor (also called the Levi-Civita symbol), \epsilon_{ijk}. You can read more at the link, but the basic idea is that \epsilon_{ijk} switches signs when two indices are swapped, so that \epsilon_{123}=-\epsilon_{213}. By convention \epsilon_{123}=+1. Necessarily \epsilon_{ijk}=0 if any two of the indices are equal, e.g. \epsilon_{112}=0.

Now, here's the point: the outer product of two tensors is a tensor. So if we have two vectors V_i and W_j, the object

T_{ij}=V_i W_j.

is a tensor. If we sum any two indices of a tensor, the result is a lower-rank tensor, e.g. if T_{ij} is a rank two tensor, then T_{ii} is a tensor of rank zero, also called a scalar (this is actually the trace of T if we think of T as a matrix). A scalar has no indices left over, so it doesn't change under rotations. In particular, if T_{ij}=V_i W_j as above, then

T_{ii}=V_i W_i=V \bullet W

is the inner product (also called the scalar product or dot product) of V and W. Now, the stuff about the magnitudes and the cosine of the angle is interesting, but the real utility of the inner product is that it takes two vectors and gives us a scalar.

Likewise, it turns out that we can take the epsilon tensor and two vectors V and W and make something that is a vector product:

U_{i}=\epsilon_{ijk}V_j W_k=V \times W.

Again, the stuff about the magnitudes and the sine of the angle is interesting, but the real interesting thing is that we have created a vector from two vectors. Actually, the cross-product is what we call a "pseudo-vector" because it changes sign under spatial inversion, but that's an aspect we don't have to explore here.

Now, hopefully, from this point you can see that there are lots of ways to make products from tensors. We could, say, make a rank-four tensor from a single vector using

T_{ijkm}=V_i V_j V_k V_m.

but there's no guarantee that this will be useful! There are rules for combining epsilons with different indices, and there's this thing called the Kronecker delta which bears a striking resemblance to the identity matrix, and it turns out that the del operator is also a rank one tensor, but that discussion is for another day...

BBB

 

[D'Alembert]

Mathematicians alway try to present mathematics as independent from experience and the nature. Einstein once raised the same question like you how mathematics can explain nature so well when it has nothing to do with it.
The clear reason for it is, that all axioms of mathematics are basing on experience. Vector algebra bases on arithmetics and arithmetics are derived from the nature.

 

[IWantToLearn]

"This geometric representation of \bf{A} \times \bf{B} is of such common occurrence that it might well be taken as the definition of the product."

He then goes on to state that "the vector product appears in mechanics in connection with couples." and later, "The product makes its appearance again in considering the velocities of the individual particles of a body which is rotating with an angular velocity given in magnitude and direction by A.

I would like to point out that in the nineteenth century, there were three other related algebraic systems being developed that became overshadowed by vector analysis.

Geometric Algebra is well worth your attention. It unifies a wide array of mathematics into one unified system including complex numbers, vector analysis, differential forms, and Pauli spinor algebra, just to mention a few. Geometric Algebra may be used in all branches of physics with some occasionally startling results.

when it comes to physics, the more tools that you have to work with the better, and Geometric Algebra, while not usually taught in universities, is a powerful tool.

Many Thanks AEM, you had enlightenment me
you had provided us with useful information, and you had opened new windows of study for me, i liked the historical analysis of the development of the subject, and i liked the idea that mathematician invent their mathematics almost inspired from nature
now i am searching for some online lectures on geometric algebra
Thanks Again

It is a projection to test for how much "one thing" is "like" "another thing."

i liked that too much

(mathematicians call a lot of things vectors that you wouldn't normally think of as vectors).

Thanks nucl34rgg,

From the standpoint of physics, what makes something a vector is how it behaves under some set of coordinate transformations. In ordinary 3d space (as opposed to 4d Minkowski space, for example), the coordinate transformations we're usually interested in are rotations, which you are invited to think of in terms of rotation matrices.

X=(1/x,1/y,1/z),

this X is not a vector. That is, 1/x_i is not a vector -- it doesn't obey the vector transformation law V'_i=R_{ij}V_j.
Now, hopefully, from this point you can see that there are lots of ways to make products from tensors. We could, say, make a rank-four tensor from a single vector using

T_{ijkm}=V_i V_j V_k V_m.

but there's no guarantee that this will be useful!

Thanks bbbeard, as you said it is too much technical, but you enlightenment me too much, specially by introducing the vector transformation law

Mathematicians alway try to present mathematics as independent from experience and the nature. Einstein once raised the same question like you how mathematics can explain nature so well when it has nothing to do with it.
The clear reason for it is, that all axioms of mathematics are basing on experience. Vector algebra bases on arithmetics and arithmetics are derived from the nature.

Thanks D'Alembert
i agree with you 100%