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Cross Product

  1. Jan 31, 2013 #1
    What is the cross product of a constant and a vector? I know that the cross product between two vectors is the area of the parallelogram those two vectors form. My intuition tells me that since a constant is not a vector, it would only be multiplying with a vector when in a cross product with one. Since the vector will only grow larger in magnitude, there would be zero area in the paralleogram formed because there is no paralleogram.
  2. jcsd
  3. Jan 31, 2013 #2
    The cross product is only defined between vectors of [itex]\mathbb{R}^3[/itex]. The cross of a constant and a vector is not defined.

  4. Jan 31, 2013 #3
    So if I had an equation that contains a term that has a cross product of a constant and a vector, do I just cross it out of the equation? ( it is in an adding term so crossing it out would be okay). That's an awesome joke(:
  5. Jan 31, 2013 #4
    Can you give a specific example?
  6. Jan 31, 2013 #5
    Sure! An equation like F=π[hXh+cXh] where h is a vector and c is a constant.
    Last edited: Jan 31, 2013
  7. Jan 31, 2013 #6
    That doesn't really make any sense.
  8. Jan 31, 2013 #7
    F is a vector.
  9. Jan 31, 2013 #8
    F=π[hXh+cXh] Sorry about not adding the equality.
  10. Jan 31, 2013 #9
    Would the term containing the cross product of the constant c and vector h in the above equation just be zero? Or am I able to take cross it out of the above equation?
  11. Jan 31, 2013 #10
    No. As it stands, your equation makes no sense. You can't take the cross product of a scalar and a vector.
  12. Jan 31, 2013 #11
    Damn that stinks. Even if the c was a constant?
  13. Jan 31, 2013 #12
    Does this equation appear in some book or anything? Can you provide some more context?
  14. Jan 31, 2013 #13
    Well I made it up haha. Im sorry. I'm new at this. Do you think you can make an equation that makes sense? Like the one I attempted but failed at.
  15. Jan 31, 2013 #14
    It only makes sense if you take the cross of a vector and a vector.

    What were you attempting to do?? What lead you to this particular equation?
  16. Jan 31, 2013 #15
    Well, the h is a vector that represents a magnetic field strength. In the definition of a current, I=dq/dt, multiplying both sides by a small length ds would give the magnetic field produced my a moving charge. (dq/dt)ds turns into dq(ds/dt) which turns into vdq where dq is a small piece of charge and v is the velocity of the total charge. Integrating both sides to I ds=vdq would give the total magnetic field. For a constant velocity, the right side of the above equation turns into vq+ c, where c is some constant. Now I get the equation h=vq+c. Solving for qv gives me h-c=qv. In the equation for magnetic force on a moving charge, F=qvxB. I substituted h-c for qv in the above force equation. B turns into uh where u is the permeability of free space. I substitute uh for B in the magnetic force equation and get F=u[hxh-cxh]. I want the cxh term to go away.
  17. Jan 31, 2013 #16
    Does that sort of help?
  18. Jan 31, 2013 #17
    I don't understand any of what you said, but my physics is very bad. I'll move this to the physics section for you.
  19. Jan 31, 2013 #18
    Thank you very much!(:
  20. Jan 31, 2013 #19


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    Science Advisor

    Saying that c is a 'constant' doesn't mean it is not a vector. A "constant" is simply something that does not change as some variable, perhaps time or a space variable, changes. In your formua c is a constant vector.
  21. Jan 31, 2013 #20
    Ohhh. That makes a lot of sense! Is there anyway I could determine what the constant vector is?
  22. Feb 2, 2013 #21
    A constant vector does not have to be a scalar !! A constant vector has a constant magnitude and a constant direction...
  23. Feb 2, 2013 #22


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    Staff Emeritus
    Science Advisor

    On the other hand, there is a generalization, the exterior product. The exterior product of a scalar and a vector is a vector. The exterior product of two vectors is a bivector. The exterior product of a vector with a bivector is a trivector. Etc.

    In 3D, there are three independent bivectors: [itex]B_{xy}, B_{yz}, B_{zx}[/itex]. The cross product can be thought of as the exterior product, combined with the identification of [itex]B_{xy}[/itex] with the unit vector [itex]\hat{z}[/itex], [itex]B_{yz}[/itex] with the unit vector [itex]\hat{x}[/itex], and [itex]B_{zx}[/itex] with the unit vector [itex]\hat{y}[/itex].

    Considering the result of the exterior product of two vectors to be another vector only works in 3D. In 2D, the exterior product of two vectors is a pseudo-scalar.
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