# Geometric Product Solutions: ABC Resolution

• Raparicio
In summary, the geometric product of two vectors can result in a scalar, a vector, or a trivector, depending on the properties of the vectors and their products. The multiplication of multivectors is not clearly defined and may require further clarification.
Raparicio
Does anybody knows where can I find the resolution of the geometrical product of abc?

Raparicio said:
Does anybody knows where can I find the resolution of the geometrical product of abc?
Can you write the geometric product of two vectors?

AM

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!

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!
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|$\hat c$ (the scalar product being 0):

abc = (ab)c = (0 + a^b)^c = (|a||b|$\hat c$)^c = |a||b|$\hat c$^c = |a||b||c| + $\hat c$ ^ $\hat c$ = |a||b||c|

(Outer product is the same as the cross product)

AM

Multivectors

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|$\hat c$ (the scalar product being 0):

abc = (ab)c = (0 + a^b)^c = (|a||b|$\hat c$)^c = |a||b|$\hat c$^c = |a||b||c| + $\hat c$ ^ $\hat c$ = |a||b||c|

(Outer product is the same as the cross product)

AM

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?

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?
In the case of the geometric product of two orthogonal vectors, you have no scalar component ($a\cdot b = 0$). 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

Andrew Mason said:
In the case of the geometric product of two orthogonal vectors, you have no scalar component ($a\cdot b = 0$). 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

I think it could be like this:

abc=a·b·c + a^b^c

but I'm not sure.

Raparicio said:
I think it could be like this:

abc=a·b·c + a^b^c

but I'm not sure.
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?

AM

$$\overrightarrow a \overrightarrow b \overrightarrow c = \overrightarrow a \left( {\overrightarrow b \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c } \right) + \overrightarrow a \bullet \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)$$

= a scaler times $\overrightarrow a$ + a scaler (triple product) + a vector perpendicular to $\overrightarrow a$

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.

Trivectors

OlderDan said:

$$\overrightarrow a \overrightarrow b \overrightarrow c = \overrightarrow a \left( {\overrightarrow b \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c } \right) + \overrightarrow a \bullet \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)$$

= a scaler times $\overrightarrow a$ + a scaler (triple product) + a vector perpendicular to $\overrightarrow a$

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.

It seems to be a vector + a trivector

Doesn't it?

Raparicio said:
It seems to be a vector + a trivector

Doesn't it?

I guess I'm not sure what you mean by a trivector. If you mean the term

$$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)$$

then yes, it includes a trivector

But when terms are combined, it looks to me like a scalar plus a vector, just like the two vector case, except that the vector is the sum of two orthogonal vectors. All that I have done is assume that the operation can be defined to include a vector "times" a scalar plus vector combination by treating the vector times scaler product in the traditional way.

OlderDan said:

$$\overrightarrow a \overrightarrow b \overrightarrow c = \overrightarrow a \left( {\overrightarrow b \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c } \right) + \overrightarrow a \bullet \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)$$

= a scaler times $\overrightarrow a$ + a scaler (triple product) + a vector perpendicular to $\overrightarrow a$

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.
Dan, how are you defining:

$$\vec a (\vec b \cdot \vec c)$$ ?

AM

abc-cba

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).

?

Andrew Mason said:
Dan, how are you defining:

$$\vec a (\vec b \cdot \vec c)$$ ?

AM

As the scalar result of the dot product times the vector a. I'm not sure this is the right way to do it. I am just speculating. If the geometric product applied to a vector yields a scalar, and you are going to take another geometric product, you have to do something with the scalar. This just seems like a reaonable something.

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).

?

a·(b·c)

This term cannot work, as Andrew noted earlier. (b·c) is not a vector.

OlderDan said:
a·(b·c)

This term cannot work, as Andrew noted earlier. (b·c) is not a vector.

... and like scalar x vector?

Raparicio said:
... and like scalar x vector?

I'm not sure what you are asking or saying here.

a·(b·c) has no meaning if the "·" means the dot product or inner product of two vectors. b·c is a scalar quantity. You can multiply a scalar times a vector, which is what I am speculating could be what is going on in the geometric product abc, but you cannot do a dot product of a vector and a scalar. The dot product is only defined between two vectors.

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!

Perhaps this sheds light on the subject

http://en.wikipedia.org/wiki/Geometric_algebra#Inner_and_outer_product

But I really doin't know what this means. To the unenlightened, this just seems to be reducable to an identity. I guess this is what you mean by a monster.

$$} a \bullet b = \frac{1}{2}\left( {ab + ba} \right)$$

$$a \wedge b = \frac{1}{2}\left( {ab - ba} \right)$$

$$ab = a \bullet b + a \wedge b = \frac{1}{2}\left( {ab + ba} \right) + \frac{1}{2}\left( {ab - ba} \right)$$

$$bc = b \bullet c + b \wedge c = \frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)$$

$$abc = a \bullet \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right] + a \wedge \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right]$$

$$abc = \frac{1}{2}\left\{ {a\left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right] + \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right]a} \right\}$$

$$+ \frac{1}{2}\left\{ {a\left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right] - \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right]a} \right\}$$

$$abc = \frac{1}{4}\left\{ {\left[ {a\left( {bc + cb} \right) + a\left( {bc - cb} \right)} \right] + \left[ {\left( {bc + cb} \right)a + \left( {bc - cb} \right)a} \right]} \right\}$$

$$+ \frac{1}{4}\left\{ {\left[ {a\left( {bc + cb} \right) + a\left( {bc - cb} \right)} \right] - \left[ {\left( {bc + cb} \right)a + \left( {bc - cb} \right)a} \right]} \right\}$$

I guess the bottom line is I have no idea about this stuff, so I should just be quiet.

Thanks

Thanks for all.

## What is "Geometric Product Solutions: ABC Resolution"?

"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.

## How is "Geometric Product Solutions: ABC Resolution" different from other problem-solving methods?

The ABC Resolution method differs from other methods in that it uses a unified language and notation for all geometric products, making it more efficient and intuitive. It also allows for the use of higher-dimensional elements such as bivectors and trivectors, which are not possible with other methods.

## What types of problems can be solved using "Geometric Product Solutions: ABC Resolution"?

This method can be applied to a wide range of problems, including those involving rotations, reflections, translations, and other geometric transformations. It can also be used to solve problems in mechanics, computer graphics, and other fields that involve geometric calculations.

## What are the advantages of using "Geometric Product Solutions: ABC Resolution"?

One of the main advantages of this method is its ability to handle complex geometric operations in a simple and intuitive way. It also allows for the use of multiple geometric elements in a single operation, resulting in more concise and elegant solutions. Additionally, the ABC Resolution method has been shown to be computationally efficient, making it a practical choice for solving real-world problems.

## Are there any limitations to "Geometric Product Solutions: ABC Resolution"?

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.

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