Maths - Vectors - Forum discussion Paolo

By: Nobody/Anonymous - nobody
file RE: Vector division  
2005-06-06 10:04

In this page:  
In the paragraph about "Vector Division" there's a misleading drawing. The sentence describing it is also misleading. 
Let's call C the result of AxB. Let's call B' any possible vector contained in the plane normal to C, except those parallel to A.  
It is not always true that AxB'=C. On the contrary, the sentence describing the drawing and the drawing are ambiguous and seem to suggest that AxB'=C is always true. 
Actually, AxB'=C is true only if the component of vector B' normal to A is equal to the component of vector B normal to A (let's call it Bnormal). I.e., |B| * sin(theta) must be equal to |B'| * sin(theta') = |Bnormal|, where theta and theta' are the oriented angles from A to B and from A to B'.  
Graphically, it interesting to show that all of these B' vectors have their tip on a line parallel to A and passing through the tip of B. 
Paolo DL 

By: Martin Baker - martinbakerProject Admin
file RE: Vector division  
2005-06-08 00:46

Thanks very much for this correction, I have updated the diagram and text as you suggest. 
I assume that there will also be another parallel line on the other side for negative values of A x B. 
I have also changed the notation of 'inverse using triple product identity' (I exchanged A and C) so that the notation is now more consistent across the page. 
I hope you dont mind, I have also included your message here. 

By: Nobody/Anonymous - nobody
file RE: Vector division  
2005-08-01 06:45

Thank you for considering and publishing the text of my message. I believe your drawing and text in this paragraph are perfectly clear and correct now.  
If you like, you may even publish my full name (see below). 
Paolo De Leva, IUSM of Rome, Italy

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