# Homework Help: Vector proof

1. Jun 19, 2010

### 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

2. Jun 20, 2010

### tiny-tim

Welcome to PF!

Hi cordines! Welcome to PF!
Noooo

you've ignored the hint

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

3. Jun 20, 2010

### 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) ?

4. Jun 20, 2010

### tiny-tim

Yes … replace c in i) by c x d

5. Jun 20, 2010

### 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.

6. Jun 20, 2010

### tiny-tim

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

Try again.

7. Jun 20, 2010

### 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?

8. Jun 21, 2010

### tiny-tim

hi cordines!

(just got up :zzz: …)
yes!!

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

get some sleep, then try again

9. Jun 21, 2010

### Susanne217

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

10. Jun 22, 2010

### 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)