Prove the **BAC-CAB** rule by writing out both side in component form

Griffiths 1

Pairoaj

The BAC-CAB rule says that

$$A \times (B \times C) = B(A \cdot C)-C(A \cdot B)$$

so I will just check the $\hat{x}$ component, the others are done the same way

$$A \times (B \times C) = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ A_x & A_y & A_z \\ (B_yC_z-B_zC_y) & (B_zC_x-B_xC_z) & (B_xC_y-B_yC_x) \\ \end{vmatrix}$$

$$=(A_yB_xC_y-A_yB_yC_x-A_zB_zC_x+A_zB_xC_z)\,\hat{x}$$

now check

$$B(A\cdot C)-C(A \cdot B) = [B_x(A_xC_x+A_yC_y+A_zC_z)-C_x(A_xB_x+A_yB_y+A_zB_z)]\, \hat{x} $$

$$ = (A_yB_xC_y+A_zB_xC_z-A_yB_yC_x-A_zB_zC_x)\,\hat{x}$$

and so the x-components are equal

the y-component and z-component can be checked the same way and shown to be equal.

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Asked: 1 year ago