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Raparicio
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Does anybody knows where can I find the resolution of the geometrical product of abc?
Can you write the geometric product of two vectors?Raparicio said:Does anybody knows where can I find the resolution of the geometrical product of abc?
If the geometric product is defined only for vectors and since the geometric product of two vectors is not a vector (ie it is a vector + a scalar), you cannot take the geometric product of a vector and a geometric product unless a and b are mutually perpendicular.Raparicio said:Dear Andrew,
The geometrical product of two vectors is:
ab= a.b + a^b (inner and outer product).
I don't know if triple geometrical product is
abc=a.b.c + a^b^c
Or
abc=a(b.c) + a(b^c)
but if a(b.c) is not a·(b·c), the formulae becomes a great monster, imposible to manipulate.
and more
abc is a trivector?
oh my god!
Andrew Mason said:If the geometric product is defined only for vectors and since the geometric product of two vectors is not a vector (ie it is a vector + a scalar), you cannot take the geometric product of a vector and a geometric product unless a and b are mutually perpendicular.
Now if abc are all orthogonal vectors such that a^b = |a||b|[itex]\hat c[/itex] (the scalar product being 0):
abc = (ab)c = (0 + a^b)^c = (|a||b|[itex]\hat c[/itex])^c = |a||b|[itex]\hat c[/itex]^c = |a||b||c| + [itex]\hat c[/itex] ^ [itex]\hat c[/itex] = |a||b||c|
(Outer product is the same as the cross product)
AM
In the case of the geometric product of two orthogonal vectors, you have no scalar component ([itex]a\cdot b = 0[/itex]). The outer product results in a 'bivector', or an area vector whose direction is perpendicular to the area. If that bivector is parallel to the third vector, its geometric product with the third vector results in a scalar but no bivector.Raparicio said:But we have a vector, a scalar an a bivector. We can do a geometric multiplication of a vector (scalar + bivector)? This is multiplication of multivectors, not?
Andrew Mason said:In the case of the geometric product of two orthogonal vectors, you have no scalar component ([itex]a\cdot b = 0[/itex]). The outer product results in a 'bivector', or an area vector whose direction is perpendicular to the area. If that bivector is parallel to the third vector, its geometric product with the third vector results in a scalar but no bivector.
If the first two vectors are perpendicular, you don't have to multiply a 'multivector' with another vector.
Unless multiplication of multivectors is defined in some way (I am unaware of a definiton but there may be one) I don't see how you could determine the geometric product of a multivector (ie. a scalar plus a bivector) and vector.
AM
You would have to do one operation at a time: (a·b)·c or a·(b·c). Since (b·c) is a scalar, what does it mean to take the dot product of a vector, a, with a scalar (b·c)? You would have to define that first, would you not?Raparicio said:I think it could be like this:
abc=a·b·c + a^b^c
but I'm not sure.
OlderDan said:I know nothing about this topic beyond the basic vector multiplication. From what I have read here, I'm thinking this might be
[tex]
\overrightarrow a \overrightarrow b \overrightarrow c = \overrightarrow a \left( {\overrightarrow b \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c } \right) + \overrightarrow a \bullet \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) [/tex]
= a scaler times [itex]\overrightarrow a[/itex] + a scaler (triple product) + a vector perpendicular to [itex]\overrightarrow a [/itex]
That would probably make it non-associative and non-commuative, but I have not checked. Is this a possibility? It seems like a natural extension of the two vector case.
Raparicio said:It seems to be a vector + a trivector
Doesn't it?
Dan, how are you defining:OlderDan said:I know nothing about this topic beyond the basic vector multiplication. From what I have read here, I'm thinking this might be
[tex]
\overrightarrow a \overrightarrow b \overrightarrow c = \overrightarrow a \left( {\overrightarrow b \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c } \right) + \overrightarrow a \bullet \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) [/tex]
= a scaler times [itex]\overrightarrow a[/itex] + a scaler (triple product) + a vector perpendicular to [itex]\overrightarrow a [/itex]
That would probably make it non-associative and non-commuative, but I have not checked. Is this a possibility? It seems like a natural extension of the two vector case.
Andrew Mason said:Dan, how are you defining:
[tex]\vec a (\vec b \cdot \vec c)[/tex] ?
AM
Raparicio said:It coudl be like this:
abc = a(bc)=(ab)c=a(b·c+b^c)=a(b·c)+a(b^c)=
a·(b·c)+a^(b·c)+a·(b^c)+a^(b^c).
?
OlderDan said:a·(b·c)
This term cannot work, as Andrew noted earlier. (b·c) is not a vector.
Raparicio said:... and like scalar x vector?
Raparicio said:Dear Andrew,
The geometrical product of two vectors is:
ab= a.b + a^b (inner and outer product).
I don't know if triple geometrical product is
abc=a.b.c + a^b^c
Or
abc=a(b.c) + a(b^c)
but if a(b.c) is not a·(b·c), the formulae becomes a great monster, imposible to manipulate.
and more
abc is a trivector?
oh my god!
"Geometric Product Solutions: ABC Resolution" is a mathematical method used to solve problems involving geometric products, which are operations that combine two or more geometric elements (such as vectors, matrices, or quaternions). This method was developed by Dr. Leo Dorst and is based on the principles of geometric algebra.
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While the ABC Resolution method is a powerful tool for solving geometric problems, it does have some limitations. For example, it may not be suitable for problems that involve large datasets or require numerical precision. It also requires some knowledge of geometric algebra, which may be a barrier for those who are unfamiliar with this mathematical framework.