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more vectors

Name: Anonymous 2008-11-04 17:35

just a quick one guys

how do i show for all vectors a,b,c that

(AxB)x(AxC)=[(AxB).C]A

Name: Anonymous 2008-11-04 20:41

To show (AxB)x(AxC)=[(AxB).C]A for all vectors,

You must show that (AxB)x(AxC)=[(AxB).C]A for the base case, and then use the inductive hypothesis.

Hope that answers your question!

Name: 4tran 2008-11-04 23:34

Using Einstein sum notation,
ε_ijk(ε_mniA^mBⁿ)(ε_pqjA^pC^q) =
ε_ijkε_mniε_pqjA^mBⁿA^pC^q =
-(ε_ijkε_inm)ε_pqjA^mBⁿA^pC^q =
-(δ_jnδ_km - δ_jmδ_kn)ε_pqjA^mBⁿA^pC^q

Name: 4tran 2008-11-04 23:46

(δ_jmδ_knε_pqjA^mA^pBⁿC^q) - (δ_jnδ_kmε_pqjA^mA^pBⁿC^q) =
(ε_pqmA^mA^pB^kC^q) - (ε_pqnA^kA^pBⁿC^q) =
({ε_pqmA^pC^q}A^mB^k) - ({ε_pqnA^pC^q}BⁿA^k) =

Name: 4tran 2008-11-04 23:59

-({ε_pmqA^pA^m}C^qB^k) + ({ε_pnqA^pBⁿ}C^qA^k) =
-({A x A}_qC^qB^k) + ({A x B}_qC^qA^k) =
-({0}*CB^k) + ({A x B}*CA^k) =
({A x B}*C)A^k

Name: Anonymous 2008-11-05 13:50

Einstein sum notation? LOL, it's just suffix notation, Christ. Awwhhh, shiiiit

Name: Anonymous 2008-11-05 14:29

>>4
How exactly did you get from line 1 to line 2?

more vectors

1 Name: Anonymous 2008-11-04 17:35

2 Name: Anonymous 2008-11-04 20:41

3 Name: 4tran 2008-11-04 23:34

4 Name: 4tran 2008-11-04 23:46

5 Name: 4tran 2008-11-04 23:59

6 Name: Anonymous 2008-11-05 13:50