Cross products

  • Thread starter custer
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  • #1
custer
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Say A,B and C are points on a plane. By taking the magnitude of the cross product of AB and BC gives you the area of the parallelogram. The direction of the answer will be perpendicular to both AB and BC, but what I don't understand is why we are getting the area? Is the area pointing in that perpendicular direction?
This applies same for
moment = r x F
Linear velocity of a particle rotating about an axis = omega x r
Let's look at the moment, r and F is perpendicular to each other, but why is the moment perpendicular to both r and F? Shouldn't the rotation be in the direction of a tangential force? I'm confused.. Why is a cross product being treated like a conventional multiplying sign when we are actually finding a vector perpendicular to both of the displacement vectors?
 

Answers and Replies

  • #2
Thaakisfox
263
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actually the "vector" we obtain from the vector product is not really a vector in the sense you studied. Rather it is a pseudo-vector. This means that when you do an improper rotation it won't transform like a vector.

It is actually an antisymmetric operator. But since there is a bijection between the space of N-n antisymmetric and n antisymmetric linear operators we can represent the "vector" product as a vector, but this only works in 3 and 7 dimensions. This is called the Hodge-dualism
 
  • #3
Bob S
4,662
6
If you take the Lorentz force cross product of current I and magnetic field B, the Lorentz force F = I x B points in the direction of the force. If you take the vector cross product of E and H in an electromagntidc field, the Poynting vector P = integral[ E x H ]dA points in the direction of the power flow.
 
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  • #4
Ulrich
24
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Its just a question of definition. A rotation is defined trough the rotation axis. This is why the moment is perpendicular to r and F because the system will then rotate about the axis of the moment.
 
  • #5
Cantab Morgan
261
2
Why is a cross product being treated like a conventional multiplying sign when we are actually finding a vector perpendicular to both of the displacement vectors?

Help me understand in what way we're treating the cross product like a conventional multiplying sign? Cross products are not commutative, but conventional multiplication is.

By the way, I don't think it's necessarily harmful to think of an oriented area in terms of its normal vector. When doing div grad curl and all that, some authors (like Schey) use notation that keeps the normal vector and the area (scalar) separate, but others (like Kleppner and Kolenkow) just treat areas themselves as vectors.

Thaakisfox, thanks for pointing out the Hodge-dualism. I've read that the cross product was just an accident of us living in a three dimensional world. But I didn't know the bit about 7 dimensions!

Yay, hundredth post!
 

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