BAC-CAB Identity — Definition, Formula & Examples

The BAC-CAB identity is a formula that simplifies the vector triple product

\mathbf{A} \times (\mathbf{B} \times \mathbf{C})

into a combination of dot products:

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

. The name comes from reading the letters in the result: B-A-C minus C-A-B.

For vectors $\mathbf{A}, \mathbf{B}, \mathbf{C} \in \mathbb{R}^3$ , the vector triple product satisfies $\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})$ . This identity expresses a cross product of a vector with a cross product as a linear combination of the two inner vectors, with scalar coefficients given by dot products.

Key Formula

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

Where:

$\mathbf{A}$ = The outer vector in the triple product
$\mathbf{B}$ = The first vector in the inner cross product
$\mathbf{C}$ = The second vector in the inner cross product
$\mathbf{A} \cdot \mathbf{C}$ = Dot product of A and C (a scalar)
$\mathbf{A} \cdot \mathbf{B}$ = Dot product of A and B (a scalar)

How It Works

Without this identity, computing

\mathbf{A} \times (\mathbf{B} \times \mathbf{C})

requires first evaluating the inner cross product and then crossing

\mathbf{A}

with the result — two full cross product computations. The BAC-CAB rule replaces both with two dot products and two scalar-vector multiplications, which is often faster. Note that the parentheses matter:

\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) \neq (\mathbf{A} \times \mathbf{B}) \times \mathbf{C}

in general. The result always lies in the plane spanned by

\mathbf{B}

and

\mathbf{C}

Worked Example

Problem: Compute A × (B × C) for A = (1, 2, 0), B = (0, 1, 3), and C = (2, 0, 1) using the BAC-CAB identity.

Compute A · C: Take the dot product of A and C.

\mathbf{A} \cdot \mathbf{C} = (1)(2) + (2)(0) + (0)(1) = 2

Compute A · B: Take the dot product of A and B.

\mathbf{A} \cdot \mathbf{B} = (1)(0) + (2)(1) + (0)(3) = 2

Apply the identity: Substitute into B(A · C) − C(A · B).

\mathbf{B}(2) - \mathbf{C}(2) = (0, 2, 6) - (4, 0, 2) = (-4, 2, 4)

Answer: A × (B × C) = (−4, 2, 4)

Why It Matters

This identity appears throughout electromagnetism (e.g., simplifying

\nabla \times (\nabla \times \mathbf{E})

), fluid dynamics, and proofs in differential geometry. It is also essential in linear algebra courses when deriving properties of the Levi-Civita symbol and tensor contractions.

Common Mistakes

Mistake: Forgetting that the cross product is not associative, and writing A × (B × C) = (A × B) × C.

Correction: These two expressions are generally not equal. The BAC-CAB identity applies only to A × (B × C). For (A × B) × C, use the related form: (A × B) × C = B(A · C) − A(B · C).