Solving the Vector Triple Product Equation: Deduction

[cordines]

i) Show that: a x ( b x c) + b x ( c x a) + c x (a x b ) =0
I managed to this, by expanding each term using the definition of the triple vector product i.e. a x ( b x c) = (a.c)b-(a.b)c and adding the results.

ii) and deduce that

a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } =
( a x c ) x ( b x d)

I expanded each term like i did in the first an added the results an obtained:
-(d x b)(b.a) - (d.a)(c x b) - (b x a)(c.d) - (b.c)(a x d )

Clearly the result does not agree, and I can't find any means how to simplify it. Some help anyone? Thanks

 

[tiny-tim]

Welcome to PF!

Hi cordines! Welcome to PF!

I expanded each term like i did in the first …

Noooo …

you've ignored the hint …

it says "deduce", which means that you should use i) to do it.

 

[cordines]

Actually at first that's what I did by substituting a x ( b x c) = - b x ( c x a) - c x (a x b ) into d x { a x ( b x c ) } and using the triple vector definition for the other three, however it was all but in vain.

Any other suggestions on how I can substitute i) in ii) ?

 

[tiny-tim]

Any other suggestions on how I can substitute i) in ii) ?

Yes … replace c in i) by c x d

 

[cordines]

As in...

To prove: a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } = ( a x c ) x ( b x d)

Proof:

a x ( b x c ) = ( a . c )b - (a . b)c
a x { b x ( c x d ) } = { a . (c x d) }b - (a.b)(c x d) ... [1]

b x ( c x a ) = (b . a)c - ( b . c)a
b x { c x ( d x a ) } = (b . a)(c x d) - { b. (c x d) }a ... [2]

c x ( a x b ) = ( c . b )a - (c . a)b
c x { d x ( a x b) } = { (c x d) . b }a - { (c x d) . a }b ... [3]

Adding [1], [2] and [3] we obtain 0 which verifies with the proof in i).

For the last term,

d x { a x ( b x c ) } = d x { (a . c)b - (a . b) c }
= (a . c) ( d x b) - (a . b) (d x c)

which does not agree. I think I'm missing something. Did I expand it correctly? Thanks for your patience.

 

[tiny-tim]

b x ( c x a ) = (b . a)c - ( b . c)a
b x { c x ( d x a ) } = (b . a)(c x d) - { b. (c x d) }a ... [2]

c x ( a x b ) = ( c . b )a - (c . a)b
c x { d x ( a x b) } = { (c x d) . b }a - { (c x d) . a }b ... [3]

Your [2] and [3] are completely wrong.

Try again. ​

 

[cordines]

b x ( c x d ) = (b . d)c - (b . c)d
a x { b x ( c x d ) } = (a x c)(b . d) - (a x d)(b .c) ...[1]

or should [1] stay as: a x { b x ( c x d ) } = { a . (c x d) }b - (a.b)(c x d)

c x ( d x a ) = (c . a)d - (c . d)a
b x { c x ( d x a ) } = (b x d)(c . a) - (b x a)(c . d) ...[2]

d x (a x b ) = (d . b)a - (d . a)b
c x { d x (a x b) } = ( c x a)(d . b) - (c x b)(d . a) ...[3]

a x ( b x c ) = (a . c)b - (a . b) c
d x { a x ( b x c ) } = (a . c) ( d x b) - (a . b) (d x c) ...[4]

Is it correct?

 

[tiny-tim]

hi cordines!

(just got up :zzz: …)

b x ( c x d ) = (b . d)c - (b . c)d
a x { b x ( c x d ) } = (a x c)(b . d) - (a x d)(b .c) ...[1]

or should [1] stay as: a x { b x ( c x d ) } = { a . (c x d) }b - (a.b)(c x d)

yes!

(and [2] [3] and [4] should look similar)

get some sleep, then try again ​

 

[Susanne217]

i) Show that: a x ( b x c) + b x ( c x a) + c x (a x b ) =0
I managed to this, by expanding each term using the definition of the triple vector product i.e. a x ( b x c) = (a.c)b-(a.b)c and adding the results.

ii) and deduce that

a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } =
( a x c ) x ( b x d)

I expanded each term like i did in the first an added the results an obtained:
-(d x b)(b.a) - (d.a)(c x b) - (b x a)(c.d) - (b.c)(a x d )

Clearly the result does not agree, and I can't find any means how to simplify it. Some help anyone? Thanks

The proof you are seaching are derived from the axioms of the vector space and can be found in any of Schaums compendiums!

 

[cordines]

To prove: a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } = ( a x c ) x ( b x d)

Proof:

a x { b x ( c x a ) } = {a . (c x d) }b - (a.b)(c x d)

b x { c x ( d x a ) } = b x { (c.a)d - (c.d)a }
= (c.a)(b x d) - (c.d)(b x a)

c x { d x ( a x b ) } = {c . (a x b) }d - (c.d)(a x b)

d x ( a x ( b x c ) } = d x { (a . c)b - (a . b)c }
= (a . c)(d x b) - (a . b)(d x c)

Adding gives:

{a . (c x d) }b + (c . (a x b) }d = { a x c . d }b - { a x c . b }d
= ( a x c ) x ( b x d)

Proved! Thanks

 

[Rasalhague]

a x { b x ( c x a ) } = {a . (c x d) }b - (a.b)(c x d)

Looks good, well done! I just spotted one typo in the first line of your proof: you wrote a instead of d on the left hand side.