# Homework Help: Geometric product

1. Apr 25, 2005

### Raparicio

Does anybody knows where can I find the resolution of the geometrical product of abc?

2. Apr 26, 2005

### Andrew Mason

Can you write the geometric product of two vectors?

AM

3. Apr 26, 2005

### Raparicio

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!!!

4. Apr 26, 2005

### Andrew Mason

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

5. Apr 26, 2005

### Raparicio

Multivectors

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?

6. Apr 26, 2005

### Andrew Mason

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

7. Apr 27, 2005

### Raparicio

I think it could be like this:

abc=a·b·c + a^b^c

but I'm not sure.

8. Apr 27, 2005

### Andrew Mason

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

9. Apr 28, 2005

### OlderDan

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

= 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.

10. Apr 29, 2005

### Raparicio

Trivectors

It seems to be a vector + a trivector

Doesn't it???

11. Apr 29, 2005

### OlderDan

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.

12. Apr 30, 2005

### Andrew Mason

Dan, how are you defining:

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

AM

13. Apr 30, 2005

### Raparicio

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).

???

14. Apr 30, 2005

### OlderDan

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.

15. Apr 30, 2005

### OlderDan

a·(b·c)

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

16. Apr 30, 2005

### Raparicio

... and like scalar x vector???

17. May 1, 2005

### OlderDan

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.

18. May 1, 2005

### OlderDan

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.

19. May 4, 2005

### Raparicio

Thanks

Thanks for all.