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What does cross product of vectors actually mean? 
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#1
Jul1614, 12:05 PM

P: 7

I understand that dot product of vectors means projecting one vector on to the other. But I don't understand what is the physical significance of a cross product? I have read that cross product gives the area of the parallelogram which has each of the vectors as its sides....but why do we want to find the area of the parallelogram?



#2
Jul1614, 12:11 PM

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I understand that you're looking for some kind of geometric meaning, but there isn't really one. It is just a construction that comes in handy in applications. So what you really want to know is how the cross product is used in applications. In my opinion, there is no natural significance of the cross product that is easy to understand. It is however very useful, but then you should check the specific uses. 


#3
Jul1614, 12:18 PM

P: 1,048

If you don't want to find the area of a parallelogram, then don't use the cross product...
Math is math, it need not correspond to any physical situation. The math we study often does correspond to physical situations because that is the math we find useful. The physical significance of a mathematical operation is dependent on the physical situation we are modeling or describing with that math. Cross products are a kind of measure of "difference" between two vectors (in opposition to the dot product which is a measure of the "sameness" between two vectors). With a cross product the more perpendicular your two vectors are the higher your cross product's magnitude will be. If your two vectors are parallel this measure of "difference" will be zero. If they are completely perpendicular then this measure of "difference" will be the maximum, the product of the two vector's magnitude. It will also point in the direction that is furthest (in a sense) from the two vectors. One physical situation that is described by the cross product is when you have a torque. Your force vector is applied and there is lever vector to the rotation axis. The torque is the cross product of these two vectors. If your force is parallel with your lever, there will be no torque. If your force is perpendicular to the lever then your torque will be at a maximum. In between these two extremes the torque can be many different values. The torque is a quantity that is the "difference" between the lever arm vector and force vector. It is also the value of the area of a parallelogram that is formed by the two vectors of the force and lever. 


#4
Jul1614, 12:48 PM

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What does cross product of vectors actually mean?
The magnitude of the cross product of two vectors could be interpreted as a measure of the "linear independence" of the two vectors.
In many physical situations, the cross product is an indication that... [abstractly] you are dealing with a more complicated "directed quantity" (www.google.com/search?q=cross+product+differential+form) than merely a typical [polar] vector. some possibly useful links: http://www.math.oregonstate.edu/brid...ot+cross.xhtml (with applets) http://www.math.oregonstate.edu/brid.../dot+cross.pdf http://math.stackexchange.com/questi...crossproduct 


#5
Jul1714, 02:05 AM

P: 1,411

Divergence, curl, grad, line integrals,...
Read this pdf from Cornell, the medical part of it all places one would think; dot product and cross product relate to the determination of and manipulation of a few physical calculations. While you have probable learned matrix multiplication and other matrix manipulations, such as determinants, and thought what is this all good for, it all ties together with dot and cross product with the use of, and is part of calculus. http://physiology.med.cornell.edu/pe..._grad_curl.pdf 


#6
Jul1714, 03:28 AM

P: 21

While dot product could be viewed as "how much a vector is parallell to another vector", the cross product could be viewed "how much a vector is perpendicular to another vector". To me it is easiest to understand it physically by looking at Torque. With torque you can really see that "the more perpendicular" the force is applied to a lever (and the axis of rotation), the more torque (the larger cross product magnitude) you will have. If you would apply force completely "unperpendicular" (i.e. parallell) you would have zero torque (so the cross product will be zero). Hope this helps.



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