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Nov 27, 2008, 7:43:04 PM11/27/08

to Geometric_Algebra

specifically exercise 1.3 of chapter 2 asks to solve alpha x + a x.b =

c for x, where alpha is a scalar and all the rest are vectors. The

answer is given as x = c/alpha - (c.b a)/(alpha(alpha + a.b)

i can substitute it in and see it works, but cannot see how to reach

it for myself. Any chance of some clues?

cheers.

tom

c for x, where alpha is a scalar and all the rest are vectors. The

answer is given as x = c/alpha - (c.b a)/(alpha(alpha + a.b)

i can substitute it in and see it works, but cannot see how to reach

it for myself. Any chance of some clues?

cheers.

tom

Nov 27, 2008, 8:25:53 PM11/27/08

to Geometric_Algebra

I remember having trouble with that one too originally;) Try dotting

and wedging the whole equation with the constant vectors, arriving at

expressions in terms of the constants for things like:

x . b

and

x ^ a

If that's not enough of a hint then show how far you get and where you

get stuck.

Peeter

and wedging the whole equation with the constant vectors, arriving at

expressions in terms of the constants for things like:

x . b

and

x ^ a

If that's not enough of a hint then show how far you get and where you

get stuck.

Peeter

Nov 28, 2008, 8:51:55 AM11/28/08

to Geometric_Algebra

ah ha! Thank you. Dot with b and re-arrange to give an expression for

x.b, substitute that into the original equation, rearrange a bit more

- seemed to work, but why? I'd got as far as x = (c - a x.b) / alpha

before, and thought i didn't have enough equations to be able to say

anything about x.b , but it seems i can get more information out by

operating on the whole of the original equation. Instead I'd tried all

sorts of generally applicable substitutions into the original

equation, only to fill a couple of pages with scribble.

Working through the subsequent exercises, it seems that operating on

the whole equation to learn more about it is often useful. I wonder

why the text didn't introduce this heuristic. Maybe it's considered

too obvious, but nothing in my intuition from scalar algebra led me to

it.

It feels like there is a lot of flexibility in GA, but so far it feels

like you need magic to navigate that. Does the magic feel more

systematic after a while?

Thanks,

Tom

x.b, substitute that into the original equation, rearrange a bit more

- seemed to work, but why? I'd got as far as x = (c - a x.b) / alpha

before, and thought i didn't have enough equations to be able to say

anything about x.b , but it seems i can get more information out by

operating on the whole of the original equation. Instead I'd tried all

sorts of generally applicable substitutions into the original

equation, only to fill a couple of pages with scribble.

Working through the subsequent exercises, it seems that operating on

the whole equation to learn more about it is often useful. I wonder

why the text didn't introduce this heuristic. Maybe it's considered

too obvious, but nothing in my intuition from scalar algebra led me to

it.

It feels like there is a lot of flexibility in GA, but so far it feels

like you need magic to navigate that. Does the magic feel more

systematic after a while?

Thanks,

Tom

Nov 28, 2008, 9:11:03 AM11/28/08

to Geometric_Algebra

Hi Tom,

Some motivation for this can be had by looking at how one can derive

Cramer's rule by wedging to eliminate terms.

I wrote a numerical example of such a "Cramer's rule" like reduction

here:

http://sites.google.com/site/peeterjoot/geometric-algebra/ga_wiki_cramers.pdf

which I'd first seen first here:

http://www.grassmannalgebra.info/grassmannalgebra/book/index.htm

... but found it was more natural in a GA context where you have a

division operation available directly.

As for, whether things get easier, I would say yes, especially when

you attempt to solve physical problems. Some of the proofs can be

tricky (ie: tricky to avoid going in circles in particular at least

for me).

Peeter

Some motivation for this can be had by looking at how one can derive

Cramer's rule by wedging to eliminate terms.

I wrote a numerical example of such a "Cramer's rule" like reduction

here:

http://sites.google.com/site/peeterjoot/geometric-algebra/ga_wiki_cramers.pdf

which I'd first seen first here:

http://www.grassmannalgebra.info/grassmannalgebra/book/index.htm

... but found it was more natural in a GA context where you have a

division operation available directly.

As for, whether things get easier, I would say yes, especially when

you attempt to solve physical problems. Some of the proofs can be

tricky (ie: tricky to avoid going in circles in particular at least

for me).

Peeter

Jul 16, 2018, 11:06:09 PM7/16/18

to Geometric_Algebra

I hate to re-open a very old thread, however, I am lost on the same problem. I get that using dot and wedge can be means to a solution as described by Peeter in the last post to this thread, however, I tried the OP's suggested approach by dotting the original equation with b, however, as I see it, that eliminates a term that appears in the given answer. The term that appears to vanish is (alpha + a.b).

If I dot the original equation with b, and then rearrange the get an expression for x.b, what I end up with is x.b = (c.b - ax)/alpha where I assume my understanding of b.b = 1 is correct.

When I substitute that into the original equation, what I end up with is alpha x + a((c.b - ax)/alpha) = c

Rearranging that somewhat, I get alpha x - a(ax/alpha) = c- c.ba/alpha which looks somewhat like the answer, but obviously it is not correct.

That is as far as I get even if I carry out the geometric product of aax, what I get is 2x/alpha and I am do not see where that is equivalent to the missing term.

I am lost with this. I would appreciate a hint as to what I am missing or doing incorrectly.

Thank you.

Matthew

If I dot the original equation with b, and then rearrange the get an expression for x.b, what I end up with is x.b = (c.b - ax)/alpha where I assume my understanding of b.b = 1 is correct.

When I substitute that into the original equation, what I end up with is alpha x + a((c.b - ax)/alpha) = c

Rearranging that somewhat, I get alpha x - a(ax/alpha) = c- c.ba/alpha which looks somewhat like the answer, but obviously it is not correct.

That is as far as I get even if I carry out the geometric product of aax, what I get is 2x/alpha and I am do not see where that is equivalent to the missing term.

I am lost with this. I would appreciate a hint as to what I am missing or doing incorrectly.

Thank you.

Matthew

Jul 17, 2018, 8:22:44 PM7/17/18

to Geometric_Algebra

Matthew,

Keep in mind that in the subexpression "a x.b", the "x.b" is a scalar multiplying the "a", a vector.

Manfred

Jul 18, 2018, 11:33:03 PM7/18/18

to Geometric_Algebra

Thank you, Manfred!

At least for this problem, the methodology has gelled, and I completely understand the steps. I think it is going to help me move forward.

This is the first time I have tried to solve an equation like this - it was not obvious to me. Even after reading the preceding material several times to ensure that I was not missing something, I was unclear as to whether it was a geometric product, or a scalar product - even though I read that the dot product is a grade - 1 operation. I had done a re-read, but I got off on the wrong track through a misunderstanding on the meaning of b.b and other difficulties understanding what to do. However, I found a web site that works chapter two exercises, though some steps are missing - https://rfistman.com/nfcm-exercises-ch2/ and by working through the missing steps, I finally understood how to solve the equation.

Thanks again!

Matthew

At least for this problem, the methodology has gelled, and I completely understand the steps. I think it is going to help me move forward.

This is the first time I have tried to solve an equation like this - it was not obvious to me. Even after reading the preceding material several times to ensure that I was not missing something, I was unclear as to whether it was a geometric product, or a scalar product - even though I read that the dot product is a grade - 1 operation. I had done a re-read, but I got off on the wrong track through a misunderstanding on the meaning of b.b and other difficulties understanding what to do. However, I found a web site that works chapter two exercises, though some steps are missing - https://rfistman.com/nfcm-exercises-ch2/ and by working through the missing steps, I finally understood how to solve the equation.

Thanks again!

Matthew

Jul 19, 2018, 9:05:17 AM7/19/18

to Geometric_Algebra

My observation is that there is a certain general mathematical sloppiness in GA (Hestenes was a major culprit). Consider the following caveats: (a) precedence must be explicitly understood (why drop the parentheses?) and (b) there are many extensions of the dot product, and Hestenes's version is problematic. When you see the dot being used between anything other than a pair of 1-vectors, beware! My suggestion would be to use Hestenes for historical context only; use more careful authors such as Dorst.

Jul 19, 2018, 9:10:43 AM7/19/18

to Geometric_Algebra

i disagree. i recommend reading hestenes as much as possible.

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Jul 19, 2018, 9:41:24 AM7/19/18

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As an undergraduate-level neophyte, Hestenes offered the background and context I needed to understand the role and value of GA.

I am, however, interested in more careful presentations, ones starting from low-level fundamentals and working their way up. I don't need to reach the Empyrean heights of GA... I'd simply like to be able to do some basic calculations in practical settings. Dorst? Peeter Van Joot (sp?)

Thanks

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Jul 21, 2018, 5:40:30 PM7/21/18

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I am as much a neophyte, and I have started to watch the 12-part series of lectures on GA at this youtube channel https://www.youtube.com/channel/UCL_lVXCyzqBb2Xc8CrvPENg This has helped gel the early chapters of NFCM for me, and my next videos in the series will cover Keppler's Problem. As well, the early videos cover solutions to simultaneous equations in 2D, i.e., Cramer's Rule - which demonstrate the simplicity with which GA handles them. The explanation along with the Cramer's rule, to me anyway, was excellent. To me, these are the best GA videos that I have so far viewed on youtube.

Being a neophyte, I should perhaps not say much, however, I worked through the first 8-chapters of Geometric Algebra for Computer Science. By comparison, it seems like Hestenes kept things simple in what I have read so far, and for me, simplicity equates to elegance. My preference would have been for more detail in Hestenes exposition and to have had something that at least explains the approach to solutions to vector equations.

That said, I think the rules of precedence are confusing when you are a neophyte. One thing that I think will be a good approach is that given the rules of precedence, if there are two vectors next to each other in an equation, in the absence of other terms like dot or wedge products, one can be assured that the two vectors represent a geometric product.

I think that the approach that Hestenes was taking was to introduce GA into a world where there was a dominant and different mathematical system. His approach seems to me to be that of explaining how GA was able to do all that the dominant mathematical system was capable of and, perhaps, more. To use metaphor, he was swimming against the current, and I think in what I have read so far, he does an excellent job of expounding on how GA can be used in place of the dominant mathematical system.

All the best,

Matthew

Being a neophyte, I should perhaps not say much, however, I worked through the first 8-chapters of Geometric Algebra for Computer Science. By comparison, it seems like Hestenes kept things simple in what I have read so far, and for me, simplicity equates to elegance. My preference would have been for more detail in Hestenes exposition and to have had something that at least explains the approach to solutions to vector equations.

That said, I think the rules of precedence are confusing when you are a neophyte. One thing that I think will be a good approach is that given the rules of precedence, if there are two vectors next to each other in an equation, in the absence of other terms like dot or wedge products, one can be assured that the two vectors represent a geometric product.

I think that the approach that Hestenes was taking was to introduce GA into a world where there was a dominant and different mathematical system. His approach seems to me to be that of explaining how GA was able to do all that the dominant mathematical system was capable of and, perhaps, more. To use metaphor, he was swimming against the current, and I think in what I have read so far, he does an excellent job of expounding on how GA can be used in place of the dominant mathematical system.

All the best,

Matthew

Jul 22, 2018, 11:00:31 PM7/22/18

to geometri...@googlegroups.com

Matthew, thanks for these pointers. I agree w/your closing comments, too.

Cheers,

Ross

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Jul 23, 2018, 11:46:24 AM7/23/18

to Geometric_Algebra

For a reference I prefer "Geometric Algebra for Physicists" by Doran and Lasenby. Also there are the resources by Alan MacDonald -

These is also my symbolic geometric algebra software -

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