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Cross Product 
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#1
Jan3113, 02:59 PM

P: 117

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
Jan3113, 03:02 PM

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P: 18,293

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.



#3
Jan3113, 03:47 PM

P: 117

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
Jan3113, 04:26 PM

P: 117

Sure! An equation like F=π[hXh+cXh] where h is a vector and c is a constant.



#7
Jan3113, 04:47 PM

P: 117

F is a vector.



#8
Jan3113, 04:55 PM

P: 117

F=π[hXh+cXh] Sorry about not adding the equality.



#9
Jan3113, 05:33 PM

P: 117

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?



#10
Jan3113, 05:35 PM

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P: 18,293




#11
Jan3113, 05:43 PM

P: 117

Damn that stinks. Even if the c was a constant?



#12
Jan3113, 05:52 PM

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P: 18,293




#13
Jan3113, 06:09 PM

P: 117

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.



#14
Jan3113, 06:54 PM

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P: 18,293

What were you attempting to do?? What lead you to this particular equation? 


#15
Jan3113, 07:16 PM

P: 117

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 hc=qv. In the equation for magnetic force on a moving charge, F=qvxB. I substituted hc 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[hxhcxh]. I want the cxh term to go away.



#16
Jan3113, 07:18 PM

P: 117

Does that sort of help?



#17
Jan3113, 07:26 PM

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P: 18,293

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.



#18
Jan3113, 07:27 PM

P: 117

Thank you very much!(:



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