Vector triple product: BAC-CAB rule

cross product

I am currently studying Introduction to Electrodynamics, fourth edition, by David J. Griffiths. Chapter 1.1.3 Triple Products introduces the vector triple product as follows:

(ii) Vector triple product: $\mathbf{A} \times (\mathbf{B} \times \mathbf{C})$. The vector triple product can be simplified by the so-called BAC-CAB rule:

$$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) – \mathbf{C}(\mathbf{A} \cdot \mathbf{B}). \tag{1.17}$$

Notice that

$$(\mathbf{A} \times \mathbf{B}) \times \mathbf{C} = – \mathbf{C} \times (\mathbf{A} \times \mathbf{B}) = – \mathbf{A}(\mathbf{B} \cdot \mathbf{C}) + \mathbf{B}(\mathbf{A} \cdot \mathbf{C})$$

is an entirely different vector (cross-products are not associative). All higher vector products can be similarly reduced, often by repeated application of Eq. 1.17, so it is never necessary for an expression to contain more than one cross product in any term. For instance,

$$(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D}) – (\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C});$$

$$\mathbf{A} \times [ \mathbf{B} \times (\mathbf{C} \times \mathbf{D})] = \mathbf{B}[\mathbf{A} \cdot (\mathbf{C} \times \mathbf{D})] – (\mathbf{A} \cdot \mathbf{B})(\mathbf{C} \times \mathbf{D}). \tag{1.18}$$

This all seems like total gibberish to me. For vectors $\mathbf{A}$ and $\mathbf{B}$, the expression $\mathbf{A} (\mathbf{B})$ does not make sense. Furthermore, the author claims that $(\mathbf{A} \times \mathbf{B}) \times \mathbf{C} = – \mathbf{C} \times (\mathbf{A} \times \mathbf{B}) = – \mathbf{A}(\mathbf{B} \cdot \mathbf{C}) + \mathbf{B}(\mathbf{A} \cdot \mathbf{C})$; although, it is not clear to me that this is true, nor does the author justify their claim. I do not understand what the "BAC-CAB rule" is supposed to be, nor do I understand the broader points that the author is trying to make in this section.

I would greatly appreciate it if people would please take the time to clarify this.

Best Answer

The notation can be slightly confusing. Note that in the RHS of the following equation

$$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})$$

$\mathbf{A} \cdot \mathbf{C}$ is a scalar (because dot products are scalars). This means that $\mathbf{B}(\mathbf{A} \cdot \mathbf{C})$ is just the vector $\mathbf{B}$ scaled by a real number. This operation is well defined. While the proof is slightly involved, some sanity checks can be instructive. For example, we expect that $(\mathbf{A} \times (\mathbf{B} \times \mathbf{C})) \cdot \mathbf{A} = 0$ since cross product of a vector is perpendicular to the vector itself. Indeed, taking the dot product of the RHS with $\mathbf{A}$ yields,

$$(\mathbf{B}\cdot \mathbf{A})(\mathbf{A} \cdot \mathbf{C}) - (\mathbf{A}\cdot \mathbf{C})(\mathbf{A} \cdot \mathbf{B})$$

which is clearly zero since the dot product is commutative. To convince yourself I would suggest

  • Doing more of these sanity checks on the other equations you've written
  • Evaluating both sides of these equations by hand for concrete values of $\mathbf{A}, \mathbf{B}, \mathbf{C}$

EDIT: To prove the identity

$$(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D}) - (\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C})$$

use the cyclic property of the scalar triple product

$$(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D}) = \left[(\mathbf{A} \times \mathbf{B}), \mathbf{C}, \mathbf{D}\right] = \left[\mathbf{C}, \mathbf{D}, (\mathbf{A} \times \mathbf{B})\right] = \mathbf{C}\cdot (\mathbf{D}\times(\mathbf{A} \times \mathbf{B}))$$

You can expand the vector triple product using the BAC-CAB rule to get the RHS.

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