Cross Product


by quantumfoam
Tags: cross, product
quantumfoam
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#1
Jan31-13, 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.
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micromass
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#2
Jan31-13, 03:02 PM
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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.

Quote Quote by Lame Joke
"What do you get when you cross a mountain-climber with a mosquito?"
"Nothing: you can't cross a scaler with a vector"
quantumfoam
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#3
Jan31-13, 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(:

micromass
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#4
Jan31-13, 04:03 PM
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Cross Product


Quote Quote by quantumfoam View Post
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(:
Can you give a specific example?
quantumfoam
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#5
Jan31-13, 04:26 PM
P: 117
Sure! An equation like F=π[hXh+cXh] where h is a vector and c is a constant.
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#6
Jan31-13, 04:41 PM
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Quote Quote by quantumfoam View Post
Sure! An equation like π[hXh+cXh] where h is a vector and c is a constant.
That doesn't really make any sense.
quantumfoam
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#7
Jan31-13, 04:47 PM
P: 117
F is a vector.
quantumfoam
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#8
Jan31-13, 04:55 PM
P: 117
F=π[hXh+cXh] Sorry about not adding the equality.
quantumfoam
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#9
Jan31-13, 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?
micromass
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#10
Jan31-13, 05:35 PM
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Quote Quote by quantumfoam View Post
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?
No. As it stands, your equation makes no sense. You can't take the cross product of a scalar and a vector.
quantumfoam
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#11
Jan31-13, 05:43 PM
P: 117
Damn that stinks. Even if the c was a constant?
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#12
Jan31-13, 05:52 PM
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Quote Quote by quantumfoam View Post
Damn that stinks. Even if the c was a constant?
Does this equation appear in some book or anything? Can you provide some more context?
quantumfoam
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#13
Jan31-13, 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.
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#14
Jan31-13, 06:54 PM
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Quote Quote by quantumfoam View Post
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.
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?
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#15
Jan31-13, 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 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.
quantumfoam
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#16
Jan31-13, 07:18 PM
P: 117
Does that sort of help?
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#17
Jan31-13, 07:26 PM
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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.
quantumfoam
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#18
Jan31-13, 07:27 PM
P: 117
Thank you very much!(:


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