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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.

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The Pointer
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    A good way of convincing yourself of the correctness of these statements is by proving them yourself. You are correct in saying that A(B) doesn't make sense. You are incorrect in asserting that the author has claimed that it makes sense. An inner product of the kind that the author is discussing produces a real number and we can certainly multiply vectors by real numbers. – Mousedorff May 26 '20 at 10:08
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    @AbhijeetVats I understand now. Thanks! – The Pointer May 26 '20 at 13:48
  • To remember the rule, think "outer dot product first". This works for both placements of the parentheses in the vector triple product. – Gavin R. Putland Jan 06 '24 at 00:34

2 Answers2

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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.

Gerard
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  • Thanks for clarifying. Do you happen to know how the author applied the BAC-CAB rule to $$(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D}) - (\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C})?$$ I still don't understand this part. – The Pointer May 26 '20 at 13:48
  • I believe this uses the cyclic property of the scalar triple product. – Gerard May 26 '20 at 17:41
  • I understand that we get $\left[(\mathbf{A} \times \mathbf{B}), \mathbf{C}, \mathbf{D}\right] = \left[\mathbf{C}, \mathbf{D}, (\mathbf{A} \times \mathbf{B})\right]$ by commutativity of the dot product, but how do we get that $\left[\mathbf{C}, \mathbf{D}, (\mathbf{A} \times \mathbf{B})\right] = \mathbf{C}\cdot (\mathbf{D}\times(\mathbf{A} \times \mathbf{B}))$? – The Pointer May 27 '20 at 00:55
  • From $\mathbf{C}\cdot (\mathbf{D}\times(\mathbf{A} \times \mathbf{B}))$, I get $$\begin{align} \mathbf{C}\cdot (\mathbf{D}\times(\mathbf{A} \times \mathbf{B})) &= \mathbf{D} [ (\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C} ] - [ ( \mathbf{A} \times \mathbf{B} ) \cdot \mathbf{C} ] \mathbf{D} \ \ \ \text{(By the cyclic property of the scalar triple product.)} \ &= \mathbf{C} \cdot [ \mathbf{A} \cdot (\mathbf{D} \cdot \mathbf{B})] - \mathbf{C} \cdot [ \mathbf{B} \cdot ( \mathbf{D} \cdot \mathbf{A})] \ \ \ \text{(By the BAC-CAB rule.)} \end{align}$$ [...] – The Pointer May 27 '20 at 01:12
  • [...] $$\begin{align} \mathbf{C} \cdot [ \mathbf{A} \cdot (\mathbf{D} \cdot \mathbf{B})] - \mathbf{C} \cdot [ \mathbf{B} \cdot ( \mathbf{D} \cdot \mathbf{A})] &= (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D}) - (\mathbf{B} \cdot \mathbf{C})(\mathbf{A} \cdot \mathbf{D}) \ \ \ \text{(Since dot products are commutative and associative.)} \ &= (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D}) - (\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C}) \ \ \ \text{(Since dot products are commutative and associative.)} \end{align}$$ – The Pointer May 27 '20 at 01:16
  • Does that look correct to you? – The Pointer May 27 '20 at 01:17
  • I'm not sure how you apply the BAC-CAB rule in the first equation, there are no vector triple products. Also, note that $[A, B, C]$ is defined as $A\cdot (B\times C)$. Therefore, $[C, D, A\times B] = C\cdot(D\times(A\times B))$ by definition. Also, I don't understand why you think $[(A\times B), C, D] = [C, D, A\times B]$ because of the commutativity of the dot product. It follows from the cyclic property of the scalar triple product as I mention in the answer. I encourage you to look at the link explaining the scalar triple product. – Gerard May 27 '20 at 06:12
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You are correct that $\mathbf A(\mathbf B)$ does NOT make any sense, but $\mathbf A(x)$ does, where $$ x=\mathbf B\cdot\mathbf C $$ is a scalar!

Regarding the non-associativity, a better way to ground the intution about this might be to think of the geometrical implications:

$\mathbf A\times(\mathbf B\times\mathbf C)$ is a vector perpendicular to $\mathbf A$ and at the same time perpendicular to $\mathbf B\times\mathbf C$

There is really no reason to expect that that would be equal to

$(\mathbf A\times\mathbf B)\times\mathbf C$ which is a vector perpendicular to $\mathbf C$ and at the same time perpendicular to $\mathbf A\times\mathbf B$

To fully understand how the computations behind this work, you should dive in to those as well. But this gives you a birds eye view of what is at stake here.

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