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Vector Products and Possibilities

Date: 10/09/2012 at 06:12:42
From: Cyrus
Subject: problem with vector products definitions logic

Dear Sir,

I have been studying maths for a long time, but one thing keeps puzzling
me: how do we get to define vector dot products and, more importantly,
cross products?

When solving problems, you hear, "Now we take the vector cross
product...." or, "We take the dot product...." I have always failed to see
how they get to this step without knowing the answer in advance. What are
the justifications for these definitions?

I do realize that in physics there are certain phenomena that depict these
definitions, but for me that cannot be satisfactory motivation for them.



Date: 10/09/2012 at 08:57:27
From: Doctor Fenton
Subject: Re: problem with vector products definitions logic

Hi Cyrus,

Thanks for writing to Dr. Math.  

I think most linear algebra textbooks give the standard formula for the
dot product and its use to compute the angle between vectors, but in case
you haven't seen such a derivation, check

    http://mathforum.org/library/drmath/view/53928.html 
    http://mathforum.org/library/drmath/view/69160.html 

For the cross-product, you can derive the formula this way. Suppose that 
v = <v1, v2, v3> and u = <u1, u2, u3> are two vectors in R^3, and you want
to find a vector x = <x1, x2, x3>, which is perpendicular to both. Two
vectors are perpendicular if and only if their dot product is 0, so we
must have

   u1*x1 + u2*x2 + u3*x3 = 0
   v1*x1 + v2*x2 + v3*x3 = 0

In this system of two equations, there are three unknowns (x1, x2, and
x3), while the u's and the v's are all known.

We can eliminate x3 by multiplying the first equation by v3 and the second
equation by u3, and subtracting the results:

   u1*v3*x1 + u2*v3*x2 + u3*v3*x3 = 0
   u3*v1*x1 + u3*v2*x2 + u3*v3*x3 = 0

So

   u1*v3*x1 - u3*v1*x1 + u2*v3*x2 - u3*v2*x2 = 0
     (u1*v3 - u3*v1)*x1 + (u2*v3 - u3*v2)*x2 = 0

There are infinitely many solutions to this system of equations, i.e., if
you choose any value for x1, you can solve for a corresponding value of
x2, and then you can determine x3 from either of the two original
equations.

However, the last equation has the form A*x1 + B*x2 = 0, so a simple
solution is to take x1 = B and x2 = -A, giving

   x1 = u2*v3 - u3*v2      and      x2 = -(u1*v3 - u3*v1).

If we substitute these values back into one of the original equations and
solve for x3, we get

   x3 = u1*v2-u2*v1.

These are the three components of the cross-product vector. If the result
is that the x vector is the zero vector, then either one of u or v is 0,
or one vector is a non-zero scalar multiple of the other. If x is not 0,
then it will be perpendicular to both u and v, since it solves the
original system of two equations which express that fact.

If you have any questions, please write back and I will try to explain
further.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 10/09/2012 at 11:43:14
From: Cyrus
Subject: problem with vector products definitions logic

Dear Dr. Fenton,

Thank you very much for your reply; however, it does not address my
problem.

I do know how to derive the cross product and dot product. It is about THE
VERY DEFINITION of these two -- the justification of these definitions --
wherein lies my problem. How did we get the notion of defining the vector
cross product as it is defined? Is it purely because of observation in
natural phenomena, or what?

In all the textbooks I have looked at, they start by defining the vector
product and the cross product, and go on from there to work them out BASED
on these definitions. 

I do not see where they get these definitions. 

Thank you for your consideration.



Date: 10/09/2012 at 12:16:57
From: Doctor Fenton
Subject: Re: problem with vector products definitions logic

I'm not sure what you are asking. 

Is the idea of finding a vector perpendicular to two given vectors not
sufficient motivation? There are certainly many applications where
orthogonal directions are particularly convenient for computation, to say
nothing of physical phenomena, such as the motion of a charged particle in
a magnetic field where a force acts at right angles to the field and the
velocity vector.

If you read the references on the dot product, the basic problem is to
find the angle between two vectors, and the formula for the dot product
drops out of the computation. Similarly, if you need to determine a
direction orthogonal to a given plane (or two non-collinear vectors), the
cross-product drops out.

We make definitions because they are useful, and it is useful to be able
to compute the angle between vectors, and to find a vector perpendicular
to a given pair of vectors or a plane.

What else are you looking for?

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 10/09/2012 at 12:34:27
From: Cyrus
Subject: Thank you (problem with vector products definitions logic)

Thank you for your kind answer. I do appreciate it. But it seems that I
fail to make my problem clear.

It is clear to me that from the dot product, you get the angle between the
two vectors; and from cross product, you get the vector which is
perpendicular to the plane of the two vectors.

My question is: are these definitions simply based on the observation of
physical phenomena or on something else? Could one have defined them
differently?



Date: 10/09/2012 at 13:27:19
From: Doctor Fenton
Subject: Re: Thank you (problem with vector products definitions logic)

Both are connected with the idea of Euclidean (flat) geometry and the
Pythagorean theorem that we use to measure distances. If you don't use
perpendicular coordinates, the computation of distances is much more
complicated than just the square root of the sum of the squares of the
components.

We certainly use non-orthogonal coordinates, such as polar or spherical
coordinates (since a radar, for example, measures the polar or spherical
coordinates of an object), but it is usually simpler to convert back to
orthogonal Euclidean coordinates for computing distances or angles
measured from points other than the origin.

Instead of using the square root of the sum of the squares of the
distances, we can measure distances differently, such as using the sum of
the absolute values of the differences between the components of two
points. But the standard Pythagorean formula is mathematically the most
tractable.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 10/09/2012 at 16:26:53
From: Cyrus
Subject: Thank you (problem with vector products definitions logic)

Thank you for the answer, although it seems not to have slightest
relevance to my question. Is this a computer answering machine?



Date: 10/09/2012 at 16:58:43
From: Doctor Fenton
Subject: Re: Thank you (problem with vector products definitions logic)

I'm sorry, but I don't understand what you are asking. I guess I don't
think about matters in the same way you do.

I'll see if any other math doctors can help.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 10/11/2012 at 05:52:02
From: Doctor Jacques
Subject: Re: problem with vector products definitions logic

Hi Cyrus,

I think your question is, "Are there other possible definitions for vector
products?"

The answer to that question is a big YES. I will show you below how to
define other kinds of vector products. Not all of them are interesting;
but some of them are used in many applications, although less often than
the dot and cross products.

The reason why the dot and cross products are encountered so frequently is
probably linked to their use in physics. There are also deeper, purely
mathematical, reasons, which I will discuss.

I will not enter into too many details, nor give proofs of all the claims
I will make; I will rather focus on the reasons behind the concepts.

In all generality, a vector product is an operation that takes two vectors
u and v as operands (arguments) and produces a vector w. We can see that
as a function:

   f : U x V -> W              [1]

Here, u is in U, v is in V, w is in W, and w = f(u, v). U, V, and W are
vectors spaces (possibly different, but over the same scalar field).

Unless explicitly specified otherwise, we will restrict ourselves to a
particular case:

   * U = V

   * The base field is R (the real numbers).

   * The vector spaces have finite dimension. We will let n = dim,
     U = dim V, and m = dim W.

For example, in the dot product, n can be anything, and m = 1 (i.e., the
dot product is a scalar, which is essentially the same thing as a vector
space of dimension 1). In the cross product, we have n = m = 3 (the reason
of the explicit 3 will be given below).

If we consider arbitrary functions f, there is not very much to say. To
have a more interesting structure, we add some conditions that express
that the product operation is "compatible" with the vector space
structure. Specifically, we require the function f to be bilinear, i.e.,
for all vectors a, b, c in U and for all scalars k in R, we must have:

   (1) f(a + b, c) = f(a, c) + f(b, c)

   (2) f(a, b + c) = f(a, b) + f(a, c)

   (3) f(ka, b) = f(a, kb) = kf(a, b)

Note that we can also write the product in operator notation. For example,
we can write f(a, b) = a # b, where '#' is a symbol that describes the
specific product (we write a.b for the dot product and a x b for the cross
product). Using that notation, the bilinearity axioms above can be written
as:

   (1') (a + b) # c = (a # c) + (b # c)

   (2') a # (b + c) = (a # b) + (a # c)

   (3') (ka) # b = a # (kb) = k(a # b)

(1') and (2') are very similar to the distributive law. (3') is vaguely
similar to the associative law, although it is not the same thing, because
k, a, and b do not belong to the same set: k is a scalar, and a and b are
vectors (and a # b can be a vector of another dimension). This property is
sometimes called "mixed associativity." This shows that requiring the
axioms (1) - (3) is not unreasonable.

In addition, (bi-)linearity is very important in physics. Many physical
laws are linear; this is somehow related to the fact that space and time
"look the same" everywhere.

From now on, a vector product will mean a function like [1] that meets all
the requirements above.

The case m = 1
-----
Let us consider the case where m = 1, i.e., when f(u, v) is a scalar.

Such a vector product is called a bilinear form. (The expression "scalar
product" would also be appropriate, but it is often used with a more
restricted meaning.)

The simplest possible product we can define is the trivial product: 
f(u, v) = 0 for all u and v. It is easy to see that this satisfies all the
requirements, but I think you will agree that it is not very interesting.

Another simple bilinear form is the elementary product given by:

   f_{ij}(u, v) = u[i] * v[j]

Here, i and j are *fixed* integers in the range {1, ..., n}, and u[i] is
the i-th component of the vector u. This is as simple as a non-trivial
bilinear form can get. However, such a product depends heavily on the
choice of a coordinate system.

We can next combine elementary products to get something more interesting. 

Let us define:

   f(u, v) = SUM (a[i, j]*u[i]*v[j])

Here, i and j range from 1 to n, and the a[i, j] are a set of fixed real
numbers. It is easy to check that this defines a bilinear form. If we
write A for the matrix {a[i, j]}, and if we represent u and v as column
vectors, we can write the above product as:

   f(u, v) = u' A v

Here, x' is the transpose of the vector u (i.e., u' is a row vector).

Now, the important point is that any bilinear form can be written in this
way: given a bilinear form f, you simply define a[i, j] as f(e_i, e_j),
where the {e_i} are basis vectors.

The trivial product corresponds to the matrix A = 0. The elementary
product f_{ij} corresponds to a matrix that has 1 in position [i, j] and 0
everywhere else.

Another simple case is when A is the identity matrix. In that case, we
have:

   f(u, v) = SUM (u[i]*v[i])

You will recognize this as the dot product: f(u, v) = u.v. This product is
also very simple, and it has more symmetry than the previous ones: it is
invariant under many coordinate transformations, including a permutation
of the basis vectors. This is one of the reasons why it occurs so
frequently.

Bilinear forms can have some additional properties.

A bilinear form f is symmetric if f(u, v) = f(v, u) for all vectors u and
v. This will be the case if the matrix A is symmetric; in particular, the
dot product is symmetric.

A bilinear form f is said to be positive-definite if f(u, u) > 0 for all
non-zero vectors u (the axioms imply that f(0, 0) = 0). Symmetric bilinear
forms are interesting because they allow us to define the length of a
vector: we define the length |u| of the vector u as sqrt(f(u, u)), which
is legitimate if f(u, u) >= 0. The square root is important, because it
ensures that multiplying a vector by a real k multiplies its length by
|k|. This allows us to define a geometric structure on the vector space.

The dot product is positive-definite, because u.u = SUM (u[i]^2). A very
important point is that some kind of converse is true: any real symmetric
and positive-definite bilinear form is equivalent to the dot product in a
suitable coordinate system. This universality property is probably one of
the main reasons for the importance of the dot product.

The dot product has also many physical interpretations; for example, if a
force F produces a displacement u, the work done is the dot product F.u.

There are other interesting bilinear forms that are not positive-definite.
For example, in special relativity, one deals with 4-dimensional vectors
(t, x, y, z), and the form defined by this plays a special role:

  f((t1, x1, y1, z1), (t2, x2, y2, z2)) = c^2*t1*t2 - x1*x2 - y1*y2 - z1*z2

This form is not positive-definite, because, for example, if 
u = (0, 1, 0, 0), then f(u, u) = -1.

Before talking about the cross product, we will first describe two other,
more general, vector products.

The tensor product
------------------
In the case of m = 1, we have seen that any vector product can be
constructed as a linear combination of the elementary products f_{ij}. In
the general case, we can use these elementary products in another way: we
can consider all the products u[i]*v[j] as separate coordinates of a
vector in another space W. As there are n^2 such products, the dimension
of W will be m = n^2.

This product is called the tensor product, and we write f(u, v) = u (x) v,
where the second parenthetical represents a cross within a circle. A basis
of W is the set {e_i (x) e_j}, where the e_k are basis vectors of U.

This product is, in some sense, the most general product that can be
defined on U. Any such product f: U x U -> E, where E is a real vector
space, can be defined by a normal linear map:

   g : W (= U (x) U) -> E

The map g can be described by a matrix G of dimension m by n^2, where 
m = dim E.

This universality makes the tensor product (and other generalizations)
omnipresent in many fields of mathematics and physics.

We can get other universal constructions by imposing some additional
conditions, like symmetry or antisymmetry, on the tensor product.

The wedge product
-----------------
We turn now to antisymmetric vector products, i.e., products where 
f(u, v) = -f(v, u). Note that this implies that f(u, u) = 0.

Like the tensor product, we can define a "most general" antisymmetric
vector product by taking these elements as coordinates of a vector in a
vector space W:

   w[i,j] = u[i]*v[j] - u[j]*v[i]

This product is called the wedge product, and written u /\ v, where the
slashes represent an inverted 'v.' As w[i, i] = 0 and w[i, j] = -w[j, i],
there are n(n - 1)/2 independent coordinates, and we can take W as a
vector space of dimension n(n - 1)/2. Like the tensor product, this
product is "universal" in the sense that any antisymmetric vector product
can be obtained from a linear map defined on the wedge product.

In the particular case n = 3, we have n(n - 1)/2 = 3, and W has the same
dimension as U. This suggests that we may try to associate the product 
u /\ v with a vector of U, as they have the same dimension.

It turns out that, if we impose some additional conditions (like the
behavior under changes of coordinates), there are essentially two ways to
do this, corresponding to two possible orientations of the space. The
usual cross product is simply a representation of the universal wedge
product with a particular choice of orientation (a "right-hand rule"). The
fact that this is equivalent to a universal construction is one of the
main reasons for its importance.

Note that this only works with n = 3, because only in that case have we 
n = n(n - 1)/2.

The case m = n
--------------
A vector space U with a (bilinear) vector product and range of U, itself,
is called an algebra. In that case, the vector product is an internal
operation. Depending on that product's additional properties, you can have
different types of algebras -- for example, commutative or associative.

As we have seen, if n = 3, the cross product defines an algebra on R^3;
this algebra is anti-commutative, and is not associative.

Conclusion
----------
To summarize, it is possible to define many things that can be called
vector products. The fact that the examples above are more frequently
encountered stems from the fact that these products are, in some sense,
the most general products that can be defined under some reasonable
conditions. This comes also from their importance in physics, although the
two reasons are probably related.

I hope that this sheds some light on the subject; please feel free to
write back if you want to discuss this further.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 10/14/2012 at 07:41:36
From: Cyrus
Subject: Thank you (problem with vector products definitions logic)

Dear Sir,

Thank you very much for your kind and detailed answer. I really did
appreciate it. Not being a mathematician but rather an engineer, I may not
have followed all the arguments presented. However, you resolved a very
important, nagging curiosity that I had, and that was that, certainly,
other definitions of vector product can and do exist, but more popularly
we take these two definitions simply because they fit the natural physical
laws that we observe. In effect, these definitions are merely the result 
of physical observations of nature.

Thank you again. 

Yours truly,

Cyrus
Associated Topics:
High School Linear Algebra

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