# BAC CAB Rule

1. Mar 1, 2005

### shaiqbashir

Hi!

Please can u prove the following rule!

Ax(BxC)=B(A.C)-C(A.B)

this is usually remembered as "BAC CAB Rule"

i just couldnt solve this question.

2. Mar 1, 2005

### Galileo

Tedious, but straightforward is just writing out the components.
Try verifying the equation one component a time.
For example, the x-component of A X (B X C) is:

$$A_y(B_xC_y-B_yC_x)-A_z(B_zC_x-B_xC_z)$$
, which can be rewritten as:
$$B_x(A_yC_y+A_zC_z)-C_x(A_yB_y+A_zB_z)$$

you almost have the x component of B(AC) in the left term and the x-component of C(AB) in the right term. They both miss the same term $B_x(A_xC_x)$ so you can just add it to both terms, since they will cancel. This establishes the relation for the x-component. Do the same for the other two (or try to be smart and use some symmetry argument).

3. Mar 1, 2005

### shaiqbashir

Well thanks for ur reply but can u plz tell me in some more detail so that i can understand it more properly!

Thanks!

4. Mar 1, 2005

### Galileo

I`m not sure which part is unclear to you.

What I meant was:
Ax(BxC)=B(A.C)-C(A.B)
is true if and only if all three components for the vector on the left and on the right are equal. Equating one component a time makes it less cluttery.

Let:
$$\vec A = A_x \vec i + A_y \vec j + A_z \vec k$$
$$\vec B = B_x \vec i + B_y \vec j + B_z \vec k$$
$$\vec C = C_x \vec i + C_y \vec j + C_z \vec k$$

Let D = B x C:
The expression for B x C is:
$$\vec D = (B_yC_z-B_zC_y)\vec i + (B_zC_x-B_xC_z)\vec j +(B_xC_y-B_yC_x)\vec k$$
The x-component of A x D is:
$$A_yD_z-A_yD_x$$
The components of D can be gotten from the expression from B x C, so:

$$(\vec A \times \vec D)_x = A_yD_z-A_zD_y=A_y(B_xC_y-B_yC_x)-A_z(B_zC_x-B_xC_z)$$
$$=B_x(A_yC_y+A_zC_z)-C_x(A_yB_y+A_zB_z)=$$
$$B_x(A_xC_x+A_yC_y+A_zC_z)-C_x(A_xB_x+A_yB_y+A_zB_z)=\left[\vec B (\vec A \cdot \vec C)-\vec C(\vec A \cdot \vec B)\right]_x$$

Last edited: Mar 1, 2005
5. Mar 1, 2005

### dextercioby

I think using euclidean tensors is the elegant way to do it...

Daniel.

6. Mar 2, 2005

### shaiqbashir

Thank u very much Galileo but would u please explain ur last step for me that how u convert the 2nd last step to the last step.
PLzzzzzzzzzzzzzzzzzzzz
Thanks a lot again

7. Mar 2, 2005

### HallsofIvy

Staff Emeritus
Why don't YOU try working out the calculation yourself and see what you get?