Product of scalars vs vectors

  • #1
7
0
(Scalar)·(Scalar) = Scalar
(Scalar)·(Vector) = Scalar
(Vector)·(Vector) = Scalar
(Scalar)x(Scalar) = Not valid
(Scalar)x(Vector) = Vector
(Vector)x(Vector) = Vector


Did I get them right, if not why?

Thanks
 

Answers and Replies

  • #2
stevendaryl
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By "x" do you mean the vector cross-product? And by "." do you mean the vector scalar product? If the answer is yes, then those operations are only applicable to a pair of vectors. I would say there are four common types of "multiplication" involving scalars and 3D vectors:
  1. Scalar times scalar to produce a scalar (ordinary multiplication)
  2. Scalar times vector to produce a vector (scaling a vector)
  3. Vector times vector to produce a scalar (scalar or "dot" product)
  4. Vector times vector to produce a vector ("cross" product)
 
  • #3
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5. Scalar cross scalar to denote the pair (scalar,scalar). "Not valid" is misleading here.
6. Scalar cross vector also denotes a pair and no vector. If it is supposed to be a vector, the cross has to be explicitly defined, e.g as scaling.
 
  • #4
jbriggs444
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5. Scalar cross scalar to denote the pair (scalar,scalar). "Not valid" is misleading here.
You are interpreting AxB for scalars A and B in the sense of the Cartesian product of a pair of sets? [the set of ordered pairs (a,b) where a is an element of A and b is an element of B]. That is rather a stretch since A and B are not sets and their cross product would not be an ordered pair but would, rather, be a [possibly singleton] set of ordered pairs.
 
  • #5
14,413
11,726
You are interpreting AxB for scalars A and B in the sense of the Cartesian product of a pair of sets? [the set of ordered pairs (a,b) where a is an element of A and b is an element of B]. That is rather a stretch since A and B are not sets and their cross product would not be an ordered pair but would, rather, be a [possibly singleton] set of ordered pairs.
Agreed. I simply found "not valid" a bit harsh, esp. because the word scalar has been put into brackets. Sometimes I've also read lines like ##2 \times 2 = 4##.
 
  • #6
jbriggs444
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Agreed. I simply found "not valid" a bit harsh, esp. because the word scalar has been put into brackets. Sometimes I've also read lines like ##2 \times 2 = 4##.
If one insists on intepreting ##\times## in the sense of the Cartesian product and interpreting 2 in the sense of set theory and the Von Neumann construction then it would be more correct to say that ##|2 \times 2| = 4## and that ##2 \times 2## = { (0,0), (0,1), (1,0), (1,1) }

This follows since, in the Von Neumann construction, 2 = {0,1}.

Bringing us back on topic for this thread... One should interpret the ##\times## notation according to context. In the context of ##2 \times 2## and a discussion of scalars and vectors, it cannot reasonably denote either a cross product of vectors or a Cartesian product of sets. The most reasonable interpretation would as an ordinary product of integers.
 
Last edited:
  • #7
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Thank you all for your feedback.
The comments certainly help a lot.
 
  • #8
631
132
By "x" do you mean the vector cross-product? And by "." do you mean the vector scalar product? If the answer is yes, then those operations are only applicable to a pair of vectors. I would say there are four common types of "multiplication" involving scalars and 3D vectors:
  1. Scalar times scalar to produce a scalar (ordinary multiplication)
  2. Scalar times vector to produce a vector (scaling a vector)
  3. Vector times vector to produce a scalar (scalar or "dot" product)
  4. Vector times vector to produce a vector ("cross" product)
It works for 7D vectors as well.
 
  • #9
22,089
3,296
It works for 7D vectors as well.
It works in any dimension with geometric algebra as well. Also, the 7D cross product is unsatisfactory: no Jacobi identify and it's not unique.
 

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