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Cross product question

  1. Nov 30, 2003 #1
    Why does the cross product of two vectors produce a vector which is perpendicularto the plane in which the original two lie?(whenever i go to look it up it is already assumed that it is perpendicular)
  2. jcsd
  3. Nov 30, 2003 #2
    I don't know what you mean by "why". Do you mean, how do you prove that it has this property? Or what?
  4. Nov 30, 2003 #3
    yes, how do you prove it?
  5. Nov 30, 2003 #4


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    Well, you could do
    a X b . a
    a X b . b

    and you would see that both are zero for generic a and b.
  6. Nov 30, 2003 #5
    I don't quite understand how that proves that the resulting vector is perpendicular and not at any other angle
  7. Nov 30, 2003 #6
    The dot product of two vectors is:

    [tex]\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}| \cos\theta[/tex]

    where θ is the angle between u and v. If the dot product is zero for two (nonzero) vectors u and v, then the cosine of the angle between them must be zero, which means that the angle has to be 90 or 270 degrees, i.e. they're perpendicular.
  8. Nov 30, 2003 #7
    That's true, but you are talking about dot product while my question was about cross product(vector product). Ax B =C
    If vectors A and B lie in a plane, why should the resulting vector C be perpendicular to that plane?
  9. Nov 30, 2003 #8


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    Becase [itex](\vec{a} \times \vec{b}) \cdot \vec{a} = 0 = (\vec{a} \times \vec{b}) \cdot \vec{b}[/itex].
  10. Nov 30, 2003 #9
    You asked for a proof that the cross product u x v is perpendicular to u and v. NateTG gave you one: from the definition of u x v (in terms of determinants, or whatever; pick your favorite way of computing a cross product), take the dot product of u x v with either u or v. You will find that the dot product is zero. Therefore, u x v is perpendicular to u and v.
  11. Dec 1, 2003 #10


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    What definition of cross-product are you using?

    A perfectly good definition is:
    The cross product of vector u and v is the vector with length equal to length of u times length of v time sine of the angle between u and v, perpendicular to both u and v and directed according to the "right hand rule".
  12. Dec 1, 2003 #11
    I tried taking the dot product of v x u with v as Ambitwistor and NateTG suggested and got it equal to v^(2) u sin(theta) cos(theta), and when i used the component method i still couldn't get it to zero although i probably did it wrong, can someone show me how to get it to zero( i apologize for my misunderstanding)
  13. Dec 1, 2003 #12
    [tex]\vec{u} \times \vec{v} \equiv (u_y v_z-u_z v_y)\hat{x} + (u_z v_x-u_x v_z)\hat{y} + (u_x v_y-v_x u_y)\hat{z}[/tex]


    (\vec{u} \times \vec{v})\cdot\vec{u} &= (u_y v_z-u_z v_y) u_x + (u_z v_x-u_x v_z) u_y + (u_x v_y-u_y v_x) u_z \\
    &= u_x u_y v_z - u_x u_z v_y + u_y u_z v_x - u_x u_y v_z + u_x u_z v_y - u_y u_z v_x \\
    &= 0

    and similarly for [itex](\vec{u} \times \vec{v})\cdot\vec{v}[/itex].
  14. Dec 27, 2003 #13
    What I know is multiplication of vector was developed so as to make it useful for different physical and mathematical applications. It could even have been a.b = a^(1/2) x b^(1/3)
    But this isn't useful!!

    As far as vector multiplication is concerned, we know that the resulting 'thing' should have a direction too. But what direction should it be given? So we search for a unique direction and the choice of the perpendicular direction looks natural and moreover it can be used in different applications. So there.

    Tell me if i'm wrong because i'm still searching for it's history.
  15. Jan 2, 2004 #14
    Just to muddy the waters even more, I'll add Spivak's general definition:

    If [tex]v_1, \ldots, v_{n-1} \in \mathbb{R}^n}[/tex] and [tex]\varphi[/tex] is defined by

    [tex]\varphi(w) = \det \begin{pmatrix} v_1 \\ \vdots \\ v_{n-1} \\ w \end{pmatrix}[/tex]

    then [tex]\varphi[/tex] is a 1-form over [tex]\mathbb{R}^n}[/tex] and [I assume by the Riesz Rep. Theorem] there is a unique [tex]z \in \mathbb{R}^n}[/tex] such that the inner product

    [tex]\langle w,z\rangle = \varphi(w) = \det \begin{pmatrix} v_1 \\ \vdots \\ v_{n-1} \\ w \end{pmatrix}[/tex]

    This [tex]z[/tex] is denoted [tex]v_1\times \cdots \times v_{n-1}[/tex].

    Spivak finishes "It is uncommon in mathematics to have a 'product' that depends on more than two factors. In this case of two vectors in [tex]v,w\in \mathbb{R}^3[/tex], we obtain a more conventional looking product, [tex]v\times w \in \mathbb{R}^3[/tex]. For this reason it is sometimes maintained that the cross product can be defined only in [tex]\mathbb{R}^3[/tex]."

    I only think this definition is interesting because I was not aware that the cross product could be generalized to more (or less!) than two vectors (always n-1 vectors in n dimensions) until I saw this.
  16. Jan 7, 2004 #15
    I think that what welle was trying to get is how the mathematics of it work, not just showing that it works. What might be helpful in this case is is a derivation of the cross product, provided that it isn't too complicated and doesn't require knowledge that we don't have (I believe that it originally came from quaternions somehow, but I don't know how quaternions work, either.).

    For example, the non-trig version of the dot product can be derived from the applying the cosine definition. You can also think about it in the opposite direction: You can also see that, given constant maginitueds, the more different (greater angle betwee) two vectors are, the less the dot product will be (which is related the fact that (a+c)(a-c) < a^2), which coincides with the fact that the cosine decreases as the vector are more different (have a greater angle between).

    But I'm at a loss for similar explanations for the cross product.
  17. Jan 8, 2004 #16
    Basically it would be extremely good to know how, when & who Cropped up Cross Product and for what reasons.

    Anyone who knows it???
  18. Jan 30, 2004 #17

    Here's a good explanation of the historical development of the cross product.


  19. Jan 30, 2004 #18


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    Once again, what is your definition of "cross product"??
    One definition of cross product is:
    (ai+ bj+ ck)x(ui+ vj+ wk)= (bw-cv)i-(aw-cu)j+(ab-bu)k which is what Ambitwistor was doing. Using that definition the proof is a tedious but direct calculation.

    Another definition is the one I gave before:
    The cross product of vector u and v is the vector with length equal to length of u times length of v time sine of the angle between u and v, perpendicular to both u and v and directed according to the "right hand rule".

    Because you write that, for the dot product of u with uxv, you got
    "v^(2) u sin(theta) cos(theta)" you appear to be using the second definition but that includes "perpendicular to both u and v" by definition.
  20. Jan 30, 2004 #19

    matt grime

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    Part of his problem might be that he's using theta there twice for two different angles. One of which might not be 90 the other of which must be 90, and hence the expression is zero
  21. Feb 1, 2004 #20

    I think the answer to your question about the cross product is this: The cross product is a definition. You can't prove a definition...because its a definition. Similarly you cannot prove the result of a dot product. A cross product and dot products simply describe what is happening physically.

    Most textbooks are unsettling because they do not explain "why" the cross product is defined the way it is. Nearly all of mathematics was developed to solve physical or financial problems. To be honest, I don't think any one really knows "why" accept that it became convention to "define" the cross product the way it is. You have to study the history of the topic.

    You can derive the expressions that lead to the result that is "defined" as the cross product by considering a force acting at a distance on an object (torque). When you consider the trigonometry and the definition of torque what you get is the magnitude of the resulting "vector" and define its direction (because it makes the most physical sense)to be normal to the plane containing the force and position vector.

    The resulting vector defined by the cross product is sometimes called a "pseudovector" because it is not a result of an agent such as force. For basically the same reason, the centripetal acceleration of a mass times the mass is a "pseudoforce" because it is a force that is not the result of a physical entity.

    I can send you the derivations. Let me know.

    Hope this helps,

    Last edited: Feb 1, 2004
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