Dot, Cross and Triple Product Help

[kankerfist]

I have been hitting this problem from every angle for days. I have gotten to about 1 full page of proofs and still not come close. My teacher says the answer is only a couple of lines long, so I know I am missing something. The question is:

Show that:
(A x B) . (B x C) x (C x A) = (A . B x C)^2

Where A,B,C are vectors and x means cross and . means dot.

The hint he gave us is to rename one of the cross products on the left hand side to a vector D, ie: (B x C) = D. Any help would be appreciated.

 

[pka]

First of all you need to write the problem correctly using grouping symbols: \(\displaystyle \L
\left[ {A \times B} \right] \cdot \left( {\left[ {B \times C} \right] \times \left[ {C \times A} \right]} \right) = \left( {A \cdot \left[ {B \times C} \right]} \right)^2\) .

Now as was suggested let \(\displaystyle \L
D = \left[ {B \times C} \right]\) .

Note that \(\displaystyle \L
\left( {D \times \left[ {C \times A} \right]} \right) = \left( {D \cdot A} \right)C - \left( {D \cdot C} \right)A\) .

But also note that \(\displaystyle \L
\left[ {A \times B} \right] \cdot A = 0\) will be true. WHY?

Now we also have the property: \(\displaystyle \L
A \cdot \left[ {B \times C} \right] = \left[ {A \times B} \right] \cdot C\) .

Now YOU put all that together into a proof!

 

[kankerfist]

I am familiar with those properties and I have used them to get this far, but I just dont see this path leading me anywhere near the right side of the equation. Here is my left side reduction:

(A x B) . ((B x C) x (C x A))
D = (B x C)
so:
(A x B) . (D x (C x A))
= (A x B) . ((D . A)C - (D . C)A)
= (A x B) . (((B x C) . A)C - ((B x C) . C)A)
= (A x B) . (((B x C) . A)C - ((C x B) . C)A)
= (A x B) . (((B x C) . A)C - (0)A)
= (A x B) . ((B x C) . A)C

This is already more lines than the answer, so I know I have gone wrong somewhere

 

[pka]

(A x B) . ((B x C) x (C x A))
D = (B x C)
so:
(A x B) . (D x (C x A))
= (A x B) . ((D . A)C - (D . C)A)
= ([(BXC)·A][(AXB)·C]
because (AXB)·A=0

 

[kankerfist]

Sorry, but I dont see anywhere in:
= (A x B) . ((D . A)C - (D . C)A)
Where there is a (AxB).A. I think you are doing a step in your head that I am not familiar with. Thank you for the help so far though.

 

[pka]

Sorry, but I dont see anywhere in:
= (A x B)· ((D . A)C - (D . C)A)=((D . A)(A x B)·C - (D . C)(A x B)·A).

 

[kankerfist]

Ok I got it now! Thanks a lot