Vector Products and PossibilitiesDate: 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 |
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