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When I was studying cross and dot products, I learned that the cross product of two vectors A and B is perpendicular to both A and B. But my mind is unable to understand that. Since both A and B are lying in same plane, how can their cross product be in a different plane?

I also found this concept in circular motion, where the torque is outwards from the plane of paper in which the circle has been formed. When I asked my teacher about that, he told me

This is not an actual direction, this is just a symbolization. The particle doesn't actually feel any force or such in that direction.

If this is so, then a particle should also not feel a force in a uniform magnetic field in which the particle is moving perpendicular to the field, because the force is perpendicular to both the velocity and the magnetic field.

Please help. I am confused here.

Red Act
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David Wax
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  • Related: https://physics.stackexchange.com/q/14082/2451 , https://physics.stackexchange.com/q/321540/2451 and links therein. – Qmechanic Jan 06 '19 at 15:26
  • @Qmechanic Thanks for the link but I didn't found the explanation of my answer there. Hope you understand – David Wax Jan 06 '19 at 15:28
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    This confusion illustrates the fact that the cross product is not the best mathematical construct to represent torque. The torque is best represented by the plane in which the two vector live. A bivector is a more natural way to represent that, but the cross product is simpler. The vector normal to that plane does identify the plane, so it generally works, but there are other issues with it. – garyp Jan 06 '19 at 15:45
  • Does it means that there doesn't exist such direction. – David Wax Jan 06 '19 at 15:47
  • @DavidWax Please tell me if my answer solves all your queries. – Harshit Joshi Jan 06 '19 at 15:52
  • Sorry, but I don't understand your question. What's the problem with having a third vector pointing in a different direction? – FGSUZ Jan 06 '19 at 19:34
  • Think of $\vec a \times \vec b$ not as a vector, but as an antisymmetric tensor: $\frac 1 2 [\vec a \vec b - \vec b \vec a]$ that just happens to have 3 components which all rotate just like a vector. – JEB Jan 06 '19 at 19:56
  • Your teacher's point was that the cross product of two "polar" (usual) vectors is a "pseudovector", which isn't actually representing a direction. But the magnetic field is a pseudovector, and the cross product of a polar vector with a pseudovector is a polar vector, which is why $v \times B$ is an actual direction (polar vector). See https://en.wikipedia.org/wiki/Pseudovector – Red Act Jan 06 '19 at 21:49
  • Hey my question is not that. I haven't asked about torque. Why everyone is answeeing me on torque. I want to know if the direction of resultant in cross product is actual or not – David Wax Jan 07 '19 at 03:16

3 Answers3

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Maybe you should study the exterior algebra where the torque and magnetic field are 2-blades. In three dimensions, a 2-blade can be represented as its dual vector, a vector perpendicular to the 2-blade and with the same size. This vector is what is given by the cross product.

md2perpe
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Think about using a wrench to tighten a bolt. You apply a force perpendicular to the wrench arm. Your force and the wrench arm are in the same plane. You are applying a torque equal to your force times the length of the wrench arm.

Now think about what happens to the threaded bolt. It turns and goes in a direction into or out of the plane. That’s the direction of the torque.

Hope this helps.

Bob D
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Since both A and B are lying in same plane how can their resultant be in a different plane.

Cross product is not a resultant. It is defined in a manner in which you will get a vector perpendicular to $A$ & $B$.

Why define torque in perpendicular direction when describing circular motion?

That is because it gives a sense about the direction of actual acceleration. Using the words and clockwise and anticlockwise during this is stupid because the notion changes depending on whether you view the object from the front or back.

Using torque and the right hand thumb rule, you can get the sense of motion from an direction(whether viewing from front or back).

Harshit Joshi
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  • What do you mean by this sense of motion. That's what confusing me. Is my teacher correct. If this is not actual then what what about the magnetic field case. – David Wax Jan 06 '19 at 15:44
  • Yes your teacher is correct. The sense of motion means clockwise or anticlockwise motion. – Harshit Joshi Jan 06 '19 at 15:45
  • So what about the last magnetic force case – David Wax Jan 06 '19 at 15:46
  • "a particle should also does not feel a force in a uniform magnetic field in which it is moving perpendicular because the force is perpendicular" Your statement itself mentions that it does feel a force which is perpendicular to both the velocity and magnetic field. – Harshit Joshi Jan 06 '19 at 15:48