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I was asked to prove :$A \times (B \times C)=(A\cdot C)B-(A\cdot B)C$ using vector multiplication of $3$ dimension

I chose $A=(a_x,a_y,a_z)$, $B=(b_x,b_y,b_z)$, $C=(c_x,c_y,c_z)$ and start with the LHS.

$(a_x,a_y,a_z)\times [(b_x,b_y,b_z)\times(c_x,c_y,c_z)]=$

$(a_x,a_y,a_z)\times[b_xc_y\hat Z+b_xc_z\hat Y-b_yc_x\hat Z+b_yc_z\hat X -b_zc_x\hat Y -b_zc_y\hat X]=$

$(a_x,a_y,a_z)\times[(b_xc_y-b_yc_x)\hat Z+(b_xc_z-b_zc_x)\hat Y+(b_yc_z-b_zc_y)\hat X]=$

$(a_xb_xc_y-a_xb_yc_x)\hat Y+(a_xb_xc_z-a_xb_zc_x)\hat Z-(a_yb_xc_y-a_yb_yc_x)\hat X-(a_yb_yc_z-a_yb_zc_y)\hat Z-(a_zb_xc_z-a_zb_zc_x)\hat X-(a_zb_yc_z-a_zb_zc_y)\hat Y=$

$(a_xb_xc_y-a_xb_yc_x-a_zb_yc_z+a_zb_zc_y)\hat Y+(a_xb_xc_z-a_xb_zc_x-a_yb_yc_z+a_yb_zc_y)\hat Z+ (-a_yb_xc_y+a_yb_yc_x-a_zb_xc_z+a_zb_zc_x)\hat X$

the RHS is:

$(a_xc_xb_x+a_yc_yb_x+a_zc_zb_x)\hat X+(a_xc_xb_y+a_yc_yb_y+a_zc_zb_y)\hat Y+(a_xc_xb_z+a_yc_yb_z+a_zc_zb_z)\hat Z-(a_xb_xc_x+a_yb_yc_x+a_zb_zc_x)\hat X-(a_xb_xc_y+a_yb_yc_y+a_zb_zc_y)\hat Y-(a_xb_xc_z+a_yb_yc_z+a_zb_zc_z)\hat Z=$

$(a_yc_yb_x+a_zc_zb_x-a_yb_yc_x-a_zb_zc_x)\hat X +(a_xc_xb_y+a_zc_zb_y-a_xb_xc_y-a_zb_zc_y)\hat Y (a_xc_xb_z+a_yc_yb_z-a_xb_xc_z-a_yb_yc_z)\hat Z$

How should I proceed now? the RHS is a vector of a kind $(\alpha b_x,\beta b_y, \gamma b_z)+(\alpha c_x,\beta c_y, \gamma c_z)$

gbox
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    Wow, expanding both sides. That's one way to do it... –  Oct 31 '15 at 21:43
  • @Bye_World can you see a way to go from the LHS to the RHS? – gbox Oct 31 '15 at 21:45
  • You forgot to write the $\hat X$, $\hat Y$, and $\hat Z$ on the RHS. Once you do so you collect terms, cancel, and you'll (hopefully) end up with the expression you got on the LHS. –  Oct 31 '15 at 21:47
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    Look at the terms of the LHS and RHS from your edit. They are the same except apparently you made some sign errors. So once you track those down you'll have proven that the left and right are the same and then you're done. –  Oct 31 '15 at 21:59
  • @Bye_World the RHS has minus only when $c$ is the last element, LHS has minus even when $b$ is the last element – gbox Oct 31 '15 at 22:26
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    $$(b_x,b_y,b_z)\times(c_x,c_y,c_z) = (b_yc_z-b_zc_y, b_zc_x-b_xc_z, b_xc_y-b_yc_x)$$ Compare this to what you got (in particular look at the $\hat Y$ component). You made the same mistake again the second time you took the cross product. –  Oct 31 '15 at 22:29
  • @Bye_World do you mean here? $(a_xb_xc_y-a_xb_yc_x-a_zb_yc_z+a_zb_zc_y)\hat Y$? – gbox Oct 31 '15 at 22:40
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    You're just writing down the negative of the $\hat Y$ components at each cross product. If you fix that I don't see any other mistakes. –  Oct 31 '15 at 22:43
  • I am giving up, can see why it is not coming out the same – gbox Oct 31 '15 at 22:53
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    OK. Give me a few to edit your let's-expand-everything-out approach to my answer. ;) –  Oct 31 '15 at 23:00
  • Just want to add that I also answered this question using Penrose's graphical notation here – Jackozee Hakkiuz Jun 01 '18 at 03:49

3 Answers3

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If you learn index (summation) notation, then you'll be able to prove it without writing out every term:

Consider an arbitrary $p$th coordinate of the vector $A\times (B\times C)$. Then

$$\begin{align} [A\times (B\times C)]_p &= \varepsilon_{pqr}A_q(B\times C)_r \\ &= \varepsilon_{pqr}A_q\varepsilon_{rst}B_sC_t \\ &= \varepsilon_{pqr}\varepsilon_{rst}A_qB_sC_t \\ &= \varepsilon_{rpq}\varepsilon_{rst}A_qB_sC_t \\ &= (\delta_{ps}\delta_{qt}-\delta_{pt}\delta_{qs})A_qB_sC_t \\ &= \delta_{ps}\delta_{qt}A_qB_sC_t -\delta_{pt}\delta_{qs}A_qB_sC_t \\ &= A_tB_pC_t - A_sB_sC_p \\ &= (A\cdot C)B_p - (A\cdot B)C_p \\ &= [(A\cdot C)B - (A\cdot B)C]_p\end{align}$$

Because this holds for an arbitrary $p$th coordinate of both vectors, it holds for the vectors themselves.$\ \ \ \ \ \square$

You still seem to be stuck in your derivation, so here's how to do it:

$$(a_x,a_y,a_z)\times \left[(b_x,b_y,b_z)\times(c_x,c_y,c_z)\right] = (a_x,a_y,a_z)\times (b_yc_z-b_zc_y,b_zc_x-b_xc_z,b_xc_y-b_yc_x) \\ = \left(\color{red}{a_y(b_xc_y-b_yc_x) - a_z(b_zc_x-b_xc_z)}, \color{purple}{a_z(b_yc_z-b_zc_y)-a_x(b_xc_y-b_yc_x)}, \color{blue}{a_x(b_zc_x-b_xc_z)-a_y(b_yc_z-b_zc_y)}\right)$$

If you compare this to what you got on the RHS you see that it is the same.

Your problem is just that you're writing down the negative of the middle component of each cross product. If you had just switched the signs of all the affected terms, you'd have gotten the same thing.

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    your prove relays on the identity: $\epsilon_{ijk}\epsilon_{imn}=\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km}$ which is basically the BAC-CAB rule... so... is not a huge improvement – Manuel Pena Jan 18 '19 at 12:08
  • this is called the cauchy-binet identity and it's rather trivial to see: the only way it doesnt vanish is j=m and k=n or j=n and k=m, in the first case both sides are 1, in the second both are -1. q.e.d. – peter Jun 13 '21 at 09:31
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You have expanded the left side.

Now, expand the right side. With any luck, the two expansions will agree.

A note: I would have written the terms as $(x, y, z)$ instead of using hats.

marty cohen
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You could also observe that your expressions are linear in all variables and thus a verification on the standard basis suffices. Then almost everything vanishes and you have like 1 case left with a tiny sign calculation.

peter
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