# Missing a step in the proof of the BAC-CAB rule

1. Oct 17, 2015

### JulienB

1. The problem statement, all variables and given/known data

Show that a x (b x c) = b (a.c) - c(a.b).

2. Relevant equations

Only using basic vector operations.

3. The attempt at a solution

I think I'm pretty close to the solution:

b x c involves that both b and c are orthogonal to (b x c), and d = a x (b x c) involves that d is orthogonal to (b x c). Therefore, d lies on the same plane as b and c, which means d can expressed in the form:
d = xb - yc.
If I multiply both sides of the equation by a, I get a.d = x(a.b) - y(a.c)
d = a x (b x c) involves that d is perpendicular to a, therefore a.d = 0, which leads to the equation:
x(a.b) - y(a.c) = 0
x(a.b) = y(a.c)
x = ∂a.c and y = ∂a.b
Replacing x and y in the previous expression of d, I reach that:
d = ∂(b(a.c) - c(a.b))

...which is almost what I want to prove, except that I must now show that ∂ = 1. Now I know from internet research and reading on this forum that one way to obtain this result is by using vectors i,j,k instead of a,b,c. My homework also says that I can use vectors which do not alter the universality of the formula, like c = c1e1 (which is equivalent as if I would say c = c1i, if I'm not confused yet). My guess is that I have to develop both a x (b x c) and ∂(b(a.c) - c(a.b)) until coming to the same expression on both sides, right?

I feel stupid, because this step looks like the easiest. However, I still do not fully understand the basic operation of replacing vectors, and it would help me for the future if someone could clarify this for me. Here is an attempt to illustrate my confusion (let's take a x (b x c)):

b x c = (b2c3 - b3c2)e1 + (b3c1 - b1c3)e2 + (b1c2 - b2c1)e3
a
x (b x c) = (a2(b1c2 - b2c1)e3 - a3(b3c1 - b1c3)e2) + a3(b2c3 - b3c2)e1 - a1(b1c2 - b2c1)e3) + (a1(b3c1 - b1c3)e2 - a2(b2c3 - b3c2)e1)
= (a2b1c2)e3 - (a2b2c1)e3 - (a3b3c1)e2 + (a3b1c3)e2 + (a3b2c3)e1 - (a3b3c2)e1 - (a1b1c2)e3 + (a1b2c1)e3 + (a1b3c1)e2 - (a1b1c3)e2 - (a2b2c3)e1 + (a1b3c2)e1
= (a3b2c3 - a3b3c2 - a2b2c3 + a1b3c2)e1 + (a3b1c3 -a3b3c1 + a1b3c1 - a1b1c3)e2 + (a2b1c2 - a2b2c1 - a1b1c2 + a1b2c1)e3

Ehem... :P I find myself very far from proving that ∂ = 1, and I am not even certain I didn't break any vector operation rules during my development!

J.

2. Oct 17, 2015

### RUber

I would look at it this way. Start from the end and rediscover what the cross products.
Assume we are working vectors in the $\hat x, \hat y, \hat z$ basis.
$\vec A = A_x \hat x + A_y \hat y + A_z \hat z$. $\vec B$ and $\vec C$ can be similarly defined.
$\vec B ( \vec A \cdot \vec C ) - \vec C(\vec A \cdot \vec B) = \\ \qquad \qquad \left( [A_xB_xC_x +A_yB_xC_y+A_zB_xC_z]-[A_xB_xC_x +A_yB_yC_x+A_zB_zC_x]\right)\hat x \\ \qquad \qquad \left( [A_xB_yC_x +A_yB_yC_y+A_zB_yC_z]-[A_xB_xC_y +A_yB_yC_y+A_zB_zC_y]\right)\hat y \\ \qquad \qquad \left( [A_xB_zC_x +A_yB_zC_y+A_zB_zC_z]-[A_xB_xC_z +A_yB_yC_z+A_zB_zC_z]\right)\hat z$
Which you will notice has once set of canceling terms in each row.
What you are left with is:
$\left( [A_yB_xC_y+A_zB_xC_z]-[A_yB_yC_x+A_zB_zC_x]\right)\hat x \\ \left( [A_xB_yC_x +A_zB_yC_z]-[A_xB_xC_y +A_zB_zC_y]\right)\hat y \\ \left( [A_xB_zC_x +A_yB_zC_y]-[A_xB_xC_z +A_yB_yC_z]\right)\hat z$
Which can be rewritten as:
$\left( A_y [ B_xC_y-yB_yC_x]+A_z[B_xC_z-B_zC_x]\right)\hat x \\ \left( A_x[ B_yC_x -B_xC_y] +A_z[ B_yC_z-B_zC_y]\right)\hat y \\ \left( A_x [B_zC_x-B_xC_z] +A_y[B_zC_y-B_yC_z] \right)\hat z$
From there, it should only be one or two more steps to see that you have a $\vec A\times(\vec B \times \vec C)$ right in front of you.

3. Oct 17, 2015

### JulienB

Which then can again be rewritten again as:
$\left( A_y [ B_xC_y-yB_yC_x]-A_z[B_zC_x-B_xC_z]\right)\hat x \\ \left( A_z[ B_yC_z -B_zC_y] -A_x[ B_xC_y-B_yC_x]\right)\hat y \\ \left( A_x [B_zC_x-B_xC_z] -A_y[B_yC_z-B_zC_y] \right)\hat z$
...which is the development of $\vec A\times(\vec B \times \vec C)$. That's great, thanks a lot!

J.