# Newbie question on some exercise

40 views

### Frédéric de Montmollin

Feb 28, 2021, 7:50:26 PM2/28/21
to Geometric_Algebra

Hello,
I am discovering GA through New Foundations for Classical Mechanics (Hestenes).
I have difficulties playing with GA. I tried to solve some exercies, but after hours I'm still blocked.

the goal is to proove the following vector identities (p.47)
(a^b).(c^d) = b.c a.d - b.d a.c

on the othe hand

b.c a.d - b.d a.c = bcad + bcda + cbad + cbda - bdac -bdca - dbac - dbca

I try to develop using the rules for inner and outer product, but I can't find a way to go further.
It is surely very basic, but I can't find the other rules to apply. Or I may have misunderstood the whole process...
If anybody can give me a hint, it will be welcome.

### alexander arsenovic

Mar 1, 2021, 8:36:15 AM3/1/21
to Geometric_Algebra
hi
those types of problems are the hardest for me as well, as it  relies mostly and algebraic trickery.    you maybe could use the appendix on spherical trig to get a good geometrical intuition, but it wasnt obvious to me.

so
(a^b).(c^d) =  A.B,
where A and B are bivectors, so its computing the scalar part of the product of two bivectors in 3 dimensions.   A.B = <AB>0.  just to keep this in mind.
to solve, first re-arrange the first bivector,
a^b = 2ab-a.b
note
, (a.b).(c^d) =0 ,
so
(a^b).(c^d)=(ab).(c^d) = a.(b.(c^d) = a.(b.cd - b.dc)= a.d b.d - a.c b.d

the trick between 2nd and third entry is given in between 1.20 and 1.21 ( might have changed in the editions, )

hope this helps.  that book is awesome, but i also found those sections and exercises really hard. this still took me like 40min, and i have been using ga for like 5 years. !

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### Frédéric de Montmollin

Mar 1, 2021, 5:14:48 PM3/1/21
to Geometric_Algebra
Hi
Thanks a lot. It was very helpful. I got the trick finally.
I could resolve others problem that was thricky for me
Regards

### Alan Bromborsky

Mar 2, 2021, 5:59:01 PM3/2/21
to Geometric_Algebra

Attached is the output of a galgebra program that shows the sort of identities you can derive using a general metric tensor (metric tensor where all the entries are the dot products of arbitrary vectors).  The code generating the output is:

def check_generalized_BAC_CAB_formulas():
#Print_Function()
g4d = Ga('a b c d e')
(a,b,c,d,e) = g4d.mv()

gprint('g_{ij} =',g4d.g)

gprint('\\bm{a|(b*c)} =',a|(b*c))
gprint('\\bm{a|(b\\wedge c)} =',a|(b^c))
gprint('\\bm{a|(b\\wedge c\\wedge d)} =',a|(b^c^d))
gprint('\\bm{a|(b\\wedge c)+c|(a\\wedge b)+b|(c\\wedge a)} =',(a|(b^c))+(c|(a^b))+(b|(c^a)))
gprint('\\bm{a*(b\\wedge c)-b*(a\\wedge c)+c*(a\\wedge b)} =',a*(b^c)-b*(a^c)+c*(a^b))
gprint('\\bm{a*(b\\wedge c\\wedge d)-b*(a\\wedge c\\wedge d)+c*(a\\wedge b\\wedge d)-d*(a\\wedge b\\wedge c)} =',a*(b^c^d)-b*(a^c^d)+c*(a^b^d)-d*(a^b^c))
gprint('\\bm{(a\\wedge b)|(c\\wedge d)} =',(a^b)|(c^d))
gprint('\\bm{((a\\wedge b)|c)|d} =',((a^b)|c)|d)
gprint('\\bm{(a\\wedge b)\\times (c\\wedge d)} =',Ga.com(a\\wedge b,c\\wedge d))
gprint('\\bm{(a\\wedge b\\wedge c)(d\\wedge e)} =',((a^b^c)*(d^e)).Fmt(2))
return
latex_check.pdf