Vector Triple Product Identity via Component Proof

One-line statement:

a \times (b \times c) = b (a \cdot c) - c (a \cdot b)

, proven directly by expanding all three vectors into Cartesian components. Key result in words: The double cross product of three vectors collapses into a difference of two simpler vectors, each a scalar (dot product) times a vector. The component proof verifies this by brute-force algebra. Where it applies: Vector calculus, classical mechanics (centripetal/Coriolis terms), electromagnetism, fluid dynamics, differential geometry. Who/when: Identity attributed to the development of vector analysis by J. W. Gibbs and O. Heaviside in the 1880s, building on Hamilton's quaternions (1843).

The vector triple product identity (also called the "BAC–CAB rule") states that for any three vectors in $R^{3}$ , the expression $a \times (b \times c)$ equals $b (a \cdot c) - c (a \cdot b)$ . It matters because the nested cross product appears constantly in physics and geometry, and the identity replaces an awkward second cross product with two dot products — far easier to compute and interpret. The component proof establishes this identity by writing every vector in terms of its Cartesian coordinates, expanding both sides symbol by symbol, and showing they agree term-for-term. A surprising fact: the result lies entirely in the plane spanned by $b$ and $c$ , never pointing along $a$ — a geometric truth that falls out automatically from the algebra. The component proof, though tedious, is the most elementary and self-contained route, requiring nothing beyond the definitions of the dot and cross products.

The Core Insight

A cross product $b \times c$ is perpendicular to the $b$ – $c$ plane. Crossing it again with $a$ rotates the result back into that plane. Therefore $a \times (b \times c)$ must be a linear combination of $b$ and $c$ — and the only question is what the coefficients are. The component proof answers this mechanically: the coefficients turn out to be exactly $(a \cdot c)$ and $- (a \cdot b)$ . The mnemonic "BAC minus CAB" encodes the order.

Formal Statement

Dot and Cross Products in Components

Let $a = (a_{1}, a_{2}, a_{3})$ , $b = (b_{1}, b_{2}, b_{3})$ , $c = (c_{1}, c_{2}, c_{3})$ be vectors in $R^{3}$ expressed in an orthonormal Cartesian basis ${\hat{e}_{1}, \hat{e}_{2}, \hat{e}_{3}}$ . The dot product is

a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = i = 1 \sum 3 a_{i} b_{i} .

⟳Sum the products of corresponding components; orthonormality means

\hat{e}_{i} \cdot \hat{e}_{j} = δ_{ij}

The cross product has components

(b \times c)_{i} = j, k = 1 \sum 3 ε_{ij k} b_{j} c_{k},

⟳

ε_{ij k}

is the Levi-Civita symbol; expanding the determinant form

det [\hat{e}_{i}; b_{j}; c_{k}]

gives this antisymmetric sum.

where $ε_{ij k} = + 1$ for cyclic permutations $(1, 2, 3)$ , $- 1$ for anticyclic, and $0$ if any index repeats.

Vector Triple Product Identity

For all $a, b, c \in R^{3}$ ,

a \times (b \times c) = b (a \cdot c) - c (a \cdot b) .

⟳Read aloud as "BAC minus CAB." Note the inner-parenthesised vectors

b, c

become the outer vectors, scaled by dot products of the remaining pair.

The companion identity (with the parenthesis on the left) is

(a \times b) \times c = b (a \cdot c) - a (b \cdot c) .

⟳Apply [3] to

- c \times (a \times b)

using anticommutativity

x \times y = - y \times x

Why It Holds / Derivation

:::proof[Component Proof of Equation [3]] We prove the identity one Cartesian component at a time, then assemble. Work with the first component; the others follow by cyclic symmetry.

Step 1 — Inner cross product. Write

d = b \times c

with components

d_{1} = b_{2} c_{3} - b_{3} c_{2}, d_{2} = b_{3} c_{1} - b_{1} c_{3}, d_{3} = b_{1} c_{2} - b_{2} c_{1} .

⟳Specialise [2] for each free index

i

; each row picks the two non-repeating indices with the cyclic sign.

Step 2 — Outer cross product, first component. By the same cross-product rule applied to

a \times d

(a \times d)_{1} = a_{2} d_{3} - a_{3} d_{2} .

⟳Use [2] with

i = 1

: only

(j, k) = (2, 3)

and

(3, 2)

contribute, with signs

+, -

Step 3 — Substitute the inner components. Insert [5] into [6]:

(a \times d)_{1} = a_{2} (b_{1} c_{2} - b_{2} c_{1}) - a_{3} (b_{3} c_{1} - b_{1} c_{3}) .

⟳Replace

d_{3}

and

d_{2}

by their definitions verbatim, keeping the overall

a_{2} (\cdot) - a_{3} (\cdot)

structure.

Step 4 — Expand. Multiply out the four terms:

(a \times d)_{1} = a_{2} b_{1} c_{2} - a_{2} b_{2} c_{1} - a_{3} b_{3} c_{1} + a_{3} b_{1} c_{3} .

⟳Distribute

a_{2}

and

- a_{3}

over their parentheses; four monomials result.

Step 5 — The "add-and-subtract" trick. We want terms of the form

b_{1} (a \cdot c)

and

c_{1} (a \cdot b)

. The complete dot products require a missing

a_{1} b_{1} c_{1}

term. Add and subtract it:

(a \times d)_{1} = (I) a_{2} b_{1} c_{2} + a_{3} b_{1} c_{3} + a_{1} b_{1} c_{1} - (II) (a_{2} b_{2} c_{1} + a_{3} b_{3} c_{1} + a_{1} b_{1} c_{1}) .

⟳Insert

+ a_{1} b_{1} c_{1} - a_{1} b_{1} c_{1}

(which is zero) and regroup so each cluster contains all three indices for one dot product.

Step 6 — Recognise the dot products. Factor:

(I) = b_{1} (a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3}) = b_{1} (a \cdot c),

⟳

b_{1}

is common to all three terms in (I); the bracket is exactly [1] applied to

a, c

(II) = c_{1} (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}) = c_{1} (a \cdot b) .

⟳

c_{1}

is common to all three terms in (II); the bracket is [1] applied to

a, b

Step 7 — First component result. Combining [10] and [11]:

(a \times (b \times c))_{1} = b_{1} (a \cdot c) - c_{1} (a \cdot b) .

⟳This is precisely the first component of

b (a \cdot c) - c (a \cdot b)

Step 8 — Cyclic completion. The labels

1 \to 2 \to 3 \to 1

enter symmetrically through [2] and [5], so repeating Steps 2–7 with

i = 2

and

i = 3

gives

(a \times (b \times c))_{i} = b_{i} (a \cdot c) - c_{i} (a \cdot b), i = 1, 2, 3.

⟳Cycle indices

(1, 2, 3) \to (2, 3, 1) \to (3, 1, 2)

; the dot products

a \cdot c

and

a \cdot b

are scalars, hence unchanged.

Stacking the three components yields equation [3]. $■$

The Levi-Civita Shortcut

The same proof done in index notation collapses to one line. Using the contraction identity

i \sum ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l},

⟳This is the "epsilon-delta" identity; verify by checking the few non-zero index patterns.

one writes $(a \times (b \times c))_{i} = \sum ε_{ij k} a_{j} ε_{k l m} b_{l} c_{m}$ and applies [14] to recover [3] in three steps. The Cartesian-component proof above is [14] unpacked into explicit arithmetic — same content, more sweat.

Josiah Willard Gibbs | 1839–1903

The American physicist and chemist Gibbs distilled Hamilton's cumbersome quaternion algebra into the modern dot/cross product calculus during the 1880s, distributing his Elements of Vector Analysis (1881–1884) as printed lecture notes to students at Yale. The triple-product identity in its BAC–CAB form is a centrepiece of that streamlined system. Oliver Heaviside independently developed an essentially identical formalism in Britain, and the two are jointly credited with founding modern vector analysis. Gibbs valued the identity precisely because it eliminated nested cross products from electromagnetic field calculations.

Building Intuition

The triple product lies in the b–c plane

Why the result must be coplanar with b and c

b \times c

is normal to the

b

–

c

plane. Crossing any vector

a

with this normal produces a vector perpendicular to the normal — i.e., back in the plane. Since the plane is spanned by

b

and

c

, the answer must be

α b + β c

for scalars

α, β

. The identity tells us

α = a \cdot c

and

β = - (a \cdot b)

. You can even fix these coefficients without full expansion by testing the special cases

a = b

and

a = c

Worked Examples

Concrete numerical evaluation

Given:

a = (1, 2, 0)

b = (0, 1, 3)

c = (2, 0, 1)

(dimensionless). Method 1 — Direct. Inner product

b \times c

$(b \times c)_{1} = (1) (1) - (3) (0) = 1$
$(b \times c)_{2} = (3) (2) - (0) (1) = 6$
$(b \times c)_{3} = (0) (0) - (1) (2) = - 2$

So $b \times c = (1, 6, - 2)$ . Now $a \times (1, 6, - 2)$ :

comp 1: $(2) (- 2) - (0) (6) = - 4$
comp 2: $(0) (1) - (1) (- 2) = 2$
comp 3: $(1) (6) - (2) (1) = 4$

Direct result: $(- 4, 2, 4)$ .

Method 2 — BAC–CAB. Dot products:

a \cdot c = (1) (2) + (2) (0) + (0) (1) = 2

;

a \cdot b = (1) (0) + (2) (1) + (0) (3) = 2

b (a \cdot c) - c (a \cdot b) = 2 (0, 1, 3) - 2 (2, 0, 1) = (0, 2, 6) - (4, 0, 2) = (- 4, 2, 4) .

Answer:

a \times (b \times c) = (- 4, 2, 4)

— both methods agree.

Physics: centripetal acceleration

Given: A point at position

r

rotates with constant angular velocity

ω

. Its acceleration includes the term

a_{cp} = ω \times (ω \times r)

. Take

ω = (0, 0, 2) rad/s

and

r = (3, 0, 0) m

. Apply [3] with

a = b = ω

c = r

ω \times (ω \times r) = ω (ω \cdot r) - r (ω \cdot ω) .

Compute: $ω \cdot r = (0) (3) + (0) (0) + (2) (0) = 0$ ; $ω \cdot ω = 0 + 0 + 4 = 4 rad^{2} / s^{2}$ .

a_{cp} = 0 - (3, 0, 0) (4) = (- 12, 0, 0) m/s^{2} .

Answer:

a_{cp} = (- 12, 0, 0) m/s^{2}

— magnitude

ω^{2} r = 4 \times 3 = 12 m/s^{2}

, directed inward (toward the rotation axis), exactly as centripetal acceleration should be.

Common Misconceptions

The cross product is NOT associative

A frequent error is writing $a \times (b \times c) = (a \times b) \times c$ . They are different vectors. Compare [3] and [4]: the first gives $b (a \cdot c) - c (a \cdot b)$ , the second gives $b (a \cdot c) - a (b \cdot c)$ . The parentheses are mandatory — never omit them. The two differ by $a (b \cdot c) - c (a \cdot b)$ , which equals $b \times (c \times a)$ — encoding the Jacobi Identity.

Associativity fails explicitly

Take $a = \hat{e}_{1}$ , $b = \hat{e}_{1}$ , $c = \hat{e}_{2}$ .

$a \times (b \times c) = \hat{e}_{1} \times (\hat{e}_{1} \times \hat{e}_{2}) = \hat{e}_{1} \times \hat{e}_{3} = - \hat{e}_{2} .$
$(a \times b) \times c = (\hat{e}_{1} \times \hat{e}_{1}) \times \hat{e}_{2} = 0 \times \hat{e}_{2} = 0 .$

One gives $- \hat{e}_{2}$ , the other gives $0$ . Associativity is decisively false.

Sign and order pitfalls

The mnemonic is "BAC minus CAB," not "CAB minus BAC." The vector that survives with a plus sign is the inner-bracket vector adjacent in the dot product to $a$ — i.e., $b (a \cdot c)$ comes first. Memorise carefully: the dotted pair always involves the outer vector $a$ with the opposite inner vector. Also remember [3] only holds in $R^{3}$ — the cross product as a vector exists only in three dimensions (and, exceptionally, seven).

Connections

Where this sits in vector analysis

The component proof of the triple product is one of the foundational manipulations underpinning Vector Calculus Identities. The Levi-Civita route ([14]) connects it to Levi-Civita Symbol and Tensor Notation and the Kronecker Delta. The associativity failure highlighted above is governed by the Jacobi Identity, which makes $(R^{3}, \times)$ a Lie Algebra. The structure $b (a \cdot c) - c (a \cdot b)$ recurs in the curl-of-curl identity $\nabla \times (\nabla \times F) = \nabla (\nabla \cdot F) - \nabla^{2} F$ , central to Maxwell's Equations and Wave Equation.

Real-World Applications

Concrete uses across science

Rotating reference frames (mechanics): Centripetal acceleration $ω \times (ω \times r)$ simplifies via [3] to the familiar inward $- ω^{2} r_{⊥}$ term in gyroscope and Earth-rotation analyses.
Electromagnetic wave equation: Deriving $\nabla^{2} E = μ_{0} ε_{0} \partial_{t}^{2} E$ from Maxwell's equations requires the curl-of-curl identity, the differential analogue of [3].
Magnetohydrodynamics: The Lorentz force per unit volume $J \times B$ expanded with $B \propto \nabla \times A$ uses triple-product reductions to separate magnetic pressure and tension.
Rigid-body dynamics: Angular momentum $L = \sum m_{i} r_{i} \times (ω \times r_{i})$ expands via [3] into the inertia-tensor form $L = I ω$ .
Computer graphics: Constructing orthonormal frames and reflecting/refracting view vectors employs triple products to project onto and out of surface tangent planes.
Orbital mechanics: The Laplace–Runge–Lenz vector $A = p \times L - mk \hat{r}$ is simplified using [3] to prove its conservation in Kepler orbits.
Antenna and plasma physics: Wave-propagation tensors in anisotropic media reduce double cross products of the wavevector $k$ with the electric field $E$ .

Historical Development

From quaternions to vector calculus

1843 — W. R. Hamilton invents quaternions; products of pure quaternions encode both dot and cross products, but entangled together.
1844 — H. Grassmann publishes Die lineale Ausdehnungslehre, an abstract algebra of extension foreshadowing vector products.
1881–1884 — J. W. Gibbs distributes privately printed Elements of Vector Analysis at Yale, isolating clean dot and cross products and stating the triple-product identity.
1880s — O. Heaviside independently reformulates Maxwell's equations in vector form, using the same identities to reduce field expressions.
1901 — E. B. Wilson, drawing on Gibbs's lectures, publishes Vector Analysis, the first textbook to present the BAC–CAB rule with component proofs to students.
1900s — Tullio Levi-Civita and Gregorio Ricci-Curbastro develop tensor calculus; the permutation symbol $ε_{ij k}$ furnishes the compact epsilon-delta proof.
1920s onward — The identity becomes standard in physics curricula via texts by Abraham, Becker, and later Jackson's Classical Electrodynamics.

Subtleties & Limits

Dimensional and structural caveats

The vector triple product as written requires a genuine vector-valued cross product, which exists only in dimensions 3 and 7 (the latter via octonion multiplication). In $R^{7}$ the BAC–CAB rule fails: the seven-dimensional cross product satisfies a modified triple-product identity with extra terms, because its associated algebra lacks the same Jacobi structure. In arbitrary dimensions the natural generalisation uses the antisymmetric wedge product and the Hodge Star Operator, where " $b \times c$ " becomes $⋆ (b \land c)$ ; the triple product then reflects properties of the Clifford/exterior algebra rather than a single tidy identity. A subtle point for students: [3] is a pointwise algebraic identity, but its differential cousin (curl of curl) carries an extra subtlety — operators do not commute with the vectors they act on, so naive substitution into [3] can mislead unless one tracks the order of differentiation.