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The logic behind definitions

  1. Sep 14, 2007 #1
    one of my cousins asked me what the mathematical logic/mathematical explanation are behind the definitions of scalar product and cross product. he wants to know why the definitions are set like the way they are. i just want to make sure from you guys before i answer him. i think that there are no mathematical logic or explanations behind the definitions. they are just defined like that. am i right or wrong?

    thanks in advance.
  2. jcsd
  3. Sep 14, 2007 #2


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    You can view the scalar and cross products as mathematical definitions, in fact they are instances of a more general idea of inner and exterior (outer) products. But there is actually a geometric meaning behind them.

    Suppose I fix an origin in space (e.g. the center of the room) and consider a vector x = (x, y, z) -- that is: a point in this space. Now from Pythagoras we know that the distance from this point to the origin is [tex]\sqrt{x^2 + y^2 + z^2}[/tex]. The nice thing is, that this can also be expressed as [tex](\mathbf{x} \cdot \mathbf{x})^{1/2}[/tex], which we also call the norm [itex]||\mathbf{x}||[/itex] of x.

    Similarly, suppose that I have two vectors x and y. Taking the cross product [itex]\mathbf{x} \times \mathbf{y}[/itex] will give me a third vector which is perpendicular to both x and y. Think about the case where I have a physical problem with two clear axes (e.g. a skater going down a slope, with one axis down the slope and one perpendicular to it). Then the cross product will very easily give me a third vector to complete an orthogonal (as in, perpendicular) coordinate system.

    This is probably the most clear use of the inner and cross product, though they (and their generalizations) turn up all over the place if you do math.
  4. Sep 14, 2007 #3


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    The English word "logic" is somewhat ambiguous. You are correct if you take it literally. However, your friends probably didn't mean it literally, and rather meant something like "Why would anyone have chosen to define that?"
  5. Sep 15, 2007 #4
    no my cousin meant "why are the definitions of vector and scalar product defined like that? what is the mathematical explanation behind why they are defined like that?"

    is there really any particular reason why they are defined like that? or did the mathematicians just chose to define them that way?
  6. Sep 15, 2007 #5


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    Apparently someone decided that "scalar" needed to have more than one meanging, although there are other terms that avoid multiple meanings for "scalar".

    Scalar normally means a dimensionless number, for example kinetic energy is a scalar.

    You can find all of this information at wikipedia.



    Dot_product or inner product

    The programming language APL extends the meaning of the inner (dot) product so it could be used to multiply matrices or any size variable with any other size variable, as long as the last dimension of the left variable has the same number of elements as the first dimension of the right variable.


    Outer product and cross product not the same thing.



    Yet another inner product:
    Last edited: Sep 15, 2007
  7. Sep 15, 2007 #6


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    I don't know if my post gave the actual reason they are defined like that, but it seems at least like a very good one to me.

    Also note that the scalar product picks out the component of one vector along the other, and the cross product picks out the perpendicular one. So if you have two forces and you want to factor one in a component parallel and a component perpendicular to the other, taking the inner and cross product will do that for you.

    Again, as I already said, the scalar and cross products are specific instances of much more general concepts (namely, as Jeff Reid pointed out, inner and exterior (outer) products, respectively). So one could ask, why are those defined as they are, and is it a coincidence that these are special instances of them? (And the answer to the second question would then of course no)
  8. Sep 15, 2007 #7
    well, i understood that the cross product of two vectors gives us another vector that is perpendicular to the two vectors. but what does the dot product signify? suppose we have two vectors: i + 2j + k and 2i + j + k. now, there dot product is 5. what does this 5 mean?
  9. Sep 15, 2007 #8
    the dot product allows you to multiply a vector A with the component of another vector B that lies along A. For instance in order to find the quantity work which is force times distance, we multiply the component of force applied in the direction of the displacement with the displacement. We need this component of force instead of the entire force because it is the only part of the force contributing to work.
  10. Sep 15, 2007 #9


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    Study linear algebra, and you'll have your answer.

    The point is, an inner product space is a powerful ambient to work in. Here's just one example. If you take some vector space and some vector v in it, and, of course, if you chose a basis {b1, ..., bn} for the space, then there exist unique scalars v1, ..., vn such that v = v1 b1 + ... + vn bn. But, one needs to *find* these scalars, right?

    Well, in an inner product space (i.e. a vector space with an inner product defined on it), these scalars are known apriori, i.e. one can prove that for this same vector v, you have v = (v | b1) b1 + ... + (v | bn) bn. (Where (.|.) denotes the scalar product.) Neat, huh?

    Edit: CompuChip's post reminded me of the important fact which I forgot to point out, eg that {b1, ... ,bn} is an orthonormal basis.
    Last edited: Sep 15, 2007
  11. Sep 15, 2007 #10


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    To make it what danny said more precise: suppose you have an object which is on a flat surface. Lay the x-axis along this surface. Now suppose I pull on the object with a force F = 2i + j + k (that is, in the direction of that vector with a force equal to the magnitude of the vector which is [itex]\sqrt{6}[/itex]), the work done will only depend on the component of that force along the surface. This can be calculated by taking the inner product, and you see that - if I may put it that way - only 2N of my force is actually doing work along the forward direction and the rest does not help me pull it forward.

    To get back to your example: let me give it a "practical" use. suppose I have a coordinate system in which e1 = i + 2j + k and e2 = i + j - 2k are basis vectors. Now I can use the cross product to find a third, orthogonal basis vector, e3 = -5i + 3j - k. I can now denote any vector v by it's coordinates in this new coordinate system, namely:
    v = (v . e1) e1 + (v . e2) e2 + (v . e3) e3. In this way I can choose a coordinate system that is most convenient for a problem and express any vector in it. In fact, usually when we speak about the vector (1, -4, 3) what we mean is: the vector with those coordinates in a tacitly understood coordinate system, (1, 0, 0), (0, 1, 0) and (0, 0, 1).

    Hope that makes it more clear (not more confusing).
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