Proving bac-cab $A \times (B \times C)=(A\cdot C)B-(A\cdot B)C$

Question

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 $(\alpha b_x,\beta b_y, \gamma b_z)+(\alpha c_x,\beta c_y, \gamma c_z)$

Answer 1

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.

Answer 2

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.$\ \ \ \ \ \square$

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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.

Answer 3

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.