Proofs Chapter 1: Algebraic Identities of Dot and Cross Products

Vector Calculus Identities — Proofs

1. Algebraic Identities of Dot and Cross Products

This chapter rigorously proves the algebraic identities (1)-(12) for the dot product and the cross product on $\mathbb{R}^3$. The principal tool is component-wise computation using the Levi-Civita symbol $\varepsilon_{ijk}$ together with the Kronecker delta $\delta_{ij}$.

Notational conventions
Throughout this proof collection, $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ denote vectors in $\mathbb{R}^3$ with components $a_i, b_i, c_i, d_i$ ($i=1,2,3$). Under Einstein's summation convention, repeated indices are summed from $1$ to $3$. The Levi-Civita symbol $\varepsilon_{ijk}$ is the totally antisymmetric tensor with $\varepsilon_{123}=+1$, sign-flipped under any odd permutation, and zero whenever an index is repeated. The $\varepsilon$-$\delta$ identity $\varepsilon_{ijk}\varepsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$ is used repeatedly in this chapter.

1.1 Basic Properties

This section verifies the most fundamental properties: commutativity of the dot product, and antisymmetry and linearity of the cross product. They form the foundation for every subsequent identity.

(1) Commutativity of the dot product

$$\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \tag{1}$$
Conditions: $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$
Proof

Write the dot product in components.

\begin{equation}\mathbf{a} \cdot \mathbf{b} = \displaystyle\sum_{i=1}^{3} a_i b_i \label{eq:1-1}\end{equation}

Multiplication of real numbers is commutative, so $a_i b_i = b_i a_i$ holds for every $i$. Substitute into $\eqref{eq:1-1}$.

\begin{equation}\displaystyle\sum_{i=1}^{3} a_i b_i = \displaystyle\sum_{i=1}^{3} b_i a_i = \mathbf{b} \cdot \mathbf{a} \label{eq:1-2}\end{equation}

From $\eqref{eq:1-2}$ we conclude $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$.

Remark: The dot product is a symmetric bilinear form. By contrast, the cross product introduced below is an antisymmetric bilinear map, and the two operations exhibit complementary algebraic properties.

(2) Antisymmetry of the cross product

$$\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a} \tag{2}$$
Conditions: $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$
Proof

Express the $i$-th component of the cross product via the Levi-Civita symbol.

\begin{equation}[\mathbf{a} \times \mathbf{b}]_i = \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j b_k \label{eq:2-1}\end{equation}

$\varepsilon_{ijk}$ is antisymmetric in $j$ and $k$, that is $\varepsilon_{ikj} = -\varepsilon_{ijk}$. Swap the dummy indices $j$ and $k$ in $\eqref{eq:2-1}$.

\begin{equation}[\mathbf{a} \times \mathbf{b}]_i = \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j b_k = \displaystyle\sum_{j,k} \varepsilon_{ikj}\, a_k b_j = -\displaystyle\sum_{j,k} \varepsilon_{ijk}\, b_j a_k \label{eq:2-2}\end{equation}

The rightmost expression of $\eqref{eq:2-2}$ equals $-[\mathbf{b} \times \mathbf{a}]_i$. Since the equality holds component-wise, the vector identity $\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a}$ follows.

Remark: Setting $\mathbf{a} = \mathbf{b}$ in (2) gives $\mathbf{a} \times \mathbf{a} = -\mathbf{a} \times \mathbf{a}$, so $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ is immediate.

(3) Distributivity of the cross product

$$\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} \tag{3}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbb{R}^3$
Proof

Write the $i$-th component of the left-hand side using the Levi-Civita symbol. Summation over real numbers is linear.

\begin{equation}[\mathbf{a} \times (\mathbf{b} + \mathbf{c})]_i = \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j (b_k + c_k) \label{eq:3-1}\end{equation}

Expand $\eqref{eq:3-1}$.

\begin{equation}\displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j (b_k + c_k) = \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j b_k + \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j c_k \label{eq:3-2}\end{equation}

The right-hand side of $\eqref{eq:3-2}$ equals $[\mathbf{a} \times \mathbf{b}]_i + [\mathbf{a} \times \mathbf{c}]_i$. Component-wise equality yields the vector identity (3).

Remark: Right distributivity $(\mathbf{a} + \mathbf{b}) \times \mathbf{c} = \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c}$ is proved analogously. Alternatively, combine with (2) to get $(\mathbf{a} + \mathbf{b}) \times \mathbf{c} = -\mathbf{c} \times (\mathbf{a} + \mathbf{b}) = -(\mathbf{c} \times \mathbf{a} + \mathbf{c} \times \mathbf{b}) = \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c}$.

(4) Scalar multiplication of the cross product

$$(c\,\mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (c\,\mathbf{b}) = c\,(\mathbf{a} \times \mathbf{b}) \tag{4}$$
Conditions: $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$, $c \in \mathbb{R}$
Proof

Write the $i$-th component of the cross product via the Levi-Civita symbol and pull the scalar $c$ out of the sum.

\begin{equation}[(c\,\mathbf{a}) \times \mathbf{b}]_i = \displaystyle\sum_{j,k} \varepsilon_{ijk}\, (c\, a_j)\, b_k = c \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j b_k = c\,[\mathbf{a} \times \mathbf{b}]_i \label{eq:4-1}\end{equation}

Likewise,

\begin{equation}[\mathbf{a} \times (c\,\mathbf{b})]_i = \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j (c\, b_k) = c \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j b_k = c\,[\mathbf{a} \times \mathbf{b}]_i \label{eq:4-2}\end{equation}

From $\eqref{eq:4-1}$ and $\eqref{eq:4-2}$, the three sides agree component-wise, hence (4) holds.

Remark: Combining (3) and (4), the cross product is linear in each argument. Therefore the cross product is a bilinear map $\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$.

1.2 Scalar Triple Product

This section proves symmetry properties of the scalar value formed by combining a cross product and a dot product on three vectors—the scalar triple product (sometimes called the box product). It is characterised by a determinant representation and the sign of cyclic permutations.

(5) Cyclic symmetry of the scalar triple product

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) \tag{5}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbb{R}^3$
Proof

Express the scalar triple product with the Levi-Civita symbol.

\begin{equation}\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \displaystyle\sum_{i,j,k} a_i\, \varepsilon_{ijk}\, b_j c_k = \displaystyle\sum_{i,j,k} \varepsilon_{ijk}\, a_i b_j c_k \label{eq:5-1}\end{equation}

Equation $\eqref{eq:5-1}$ equals the determinant

\begin{equation}\det[\mathbf{a},\mathbf{b},\mathbf{c}] = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \label{eq:5-2}\end{equation}

by the Levi-Civita expansion of a determinant. Replace the indices in $\eqref{eq:5-1}$ by the cyclic permutation $(i,j,k) \to (j,k,i)$. This is an even permutation, so $\varepsilon_{ijk} = \varepsilon_{jki}$. Renaming dummy indices,

\begin{equation}\displaystyle\sum_{i,j,k} \varepsilon_{ijk}\, a_i b_j c_k = \displaystyle\sum_{i,j,k} \varepsilon_{jki}\, a_i b_j c_k = \displaystyle\sum_{i',j',k'} \varepsilon_{i'j'k'}\, b_{i'} c_{j'} a_{k'} = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) \label{eq:5-3}\end{equation}

Similarly, the even permutation $(i,j,k) \to (k,i,j)$ yields

\begin{equation}\displaystyle\sum_{i,j,k} \varepsilon_{ijk}\, a_i b_j c_k = \displaystyle\sum_{i,j,k} \varepsilon_{kij}\, a_i b_j c_k = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) \label{eq:5-4}\end{equation}

From $\eqref{eq:5-3}$ and $\eqref{eq:5-4}$, all three expressions are equal.

Remark: The scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ represents the signed volume of the parallelepiped spanned by the three vectors. It is also written $[\mathbf{a},\mathbf{b},\mathbf{c}]$ or $(\mathbf{a},\mathbf{b},\mathbf{c})$.

(6) Antisymmetry of the scalar triple product under a swap

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = -\,\mathbf{a} \cdot (\mathbf{c} \times \mathbf{b}) \tag{6}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbb{R}^3$
Proof

By antisymmetry of the cross product (2),

\begin{equation}\mathbf{b} \times \mathbf{c} = -\,\mathbf{c} \times \mathbf{b} \label{eq:6-1}\end{equation}

Take the dot product of both sides with $\mathbf{a}$. Linearity of the dot product lets us pull out the constant factor $-1$.

\begin{equation}\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{a} \cdot (-\,\mathbf{c} \times \mathbf{b}) = -\,\mathbf{a} \cdot (\mathbf{c} \times \mathbf{b}) \label{eq:6-2}\end{equation}

From $\eqref{eq:6-2}$ we obtain (6).

Remark: Combining (5) and (6) shows that $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ is invariant under even permutations of $(\mathbf{a},\mathbf{b},\mathbf{c})$ and changes sign under odd permutations—the same sign rule obeyed by a determinant.

1.3 Vector Triple Product (BAC-CAB Rule)

This section proves the BAC-CAB rule, which rewrites a doubly applied cross product as a linear combination of the original vectors weighted by dot products. The key tool is the $\varepsilon$-$\delta$ identity.

(7) BAC-CAB rule

$$\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c}) - \mathbf{c}\,(\mathbf{a} \cdot \mathbf{b}) \tag{7}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbb{R}^3$
Proof

Write the $i$-th component of the left-hand side via the Levi-Civita symbol. Setting $\mathbf{u} = \mathbf{b} \times \mathbf{c}$, we have $u_k = \displaystyle\sum_{l,m} \varepsilon_{klm}\, b_l c_m$.

\begin{equation}[\mathbf{a} \times (\mathbf{b} \times \mathbf{c})]_i = \displaystyle\sum_{j,k} \varepsilon_{ijk}\, a_j u_k = \displaystyle\sum_{j,k,l,m} \varepsilon_{ijk}\, \varepsilon_{klm}\, a_j b_l c_m \label{eq:7-1}\end{equation}

By the cyclic property $\varepsilon_{ijk} = \varepsilon_{kij}$ of the Levi-Civita symbol, $\varepsilon_{ijk} \varepsilon_{klm} = \varepsilon_{kij} \varepsilon_{klm}$. Apply the $\varepsilon$-$\delta$ identity

\begin{equation}\displaystyle\sum_{k} \varepsilon_{kij}\, \varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \label{eq:7-2}\end{equation}

to $\eqref{eq:7-1}$.

\begin{equation}[\mathbf{a} \times (\mathbf{b} \times \mathbf{c})]_i = \displaystyle\sum_{j,l,m} (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\, a_j b_l c_m \label{eq:7-3}\end{equation}

Split $\eqref{eq:7-3}$ into two terms and contract using the deltas. The first term forces $l=i, m=j$; the second forces $m=i, l=j$.

\begin{equation}\displaystyle\sum_{j,l,m} \delta_{il}\delta_{jm}\, a_j b_l c_m = \displaystyle\sum_{j} a_j b_i c_j = b_i \displaystyle\sum_{j} a_j c_j = b_i\,(\mathbf{a} \cdot \mathbf{c}) \label{eq:7-4}\end{equation}

\begin{equation}\displaystyle\sum_{j,l,m} \delta_{im}\delta_{jl}\, a_j b_l c_m = \displaystyle\sum_{j} a_j b_j c_i = c_i \displaystyle\sum_{j} a_j b_j = c_i\,(\mathbf{a} \cdot \mathbf{b}) \label{eq:7-5}\end{equation}

Substitute $\eqref{eq:7-4}$ and $\eqref{eq:7-5}$ into $\eqref{eq:7-3}$.

\begin{equation}[\mathbf{a} \times (\mathbf{b} \times \mathbf{c})]_i = b_i\,(\mathbf{a} \cdot \mathbf{c}) - c_i\,(\mathbf{a} \cdot \mathbf{b}) \label{eq:7-6}\end{equation}

Equation $\eqref{eq:7-6}$ holds component-wise, giving the vector identity $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b}(\mathbf{a} \cdot \mathbf{c}) - \mathbf{c}(\mathbf{a} \cdot \mathbf{b})$.

Remark: Reading the leading letters on the right-hand side gives "BAC minus CAB", whence the well-known mnemonic "BAC-CAB rule".

(8) Vector triple product (outer parenthesisation)

$$(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c}) - \mathbf{a}\,(\mathbf{b} \cdot \mathbf{c}) \tag{8}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbb{R}^3$
Proof

Apply antisymmetry of the cross product (2) to the left-hand side.

\begin{equation}(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = -\,\mathbf{c} \times (\mathbf{a} \times \mathbf{b}) \label{eq:8-1}\end{equation}

Apply the BAC-CAB rule (7) to $\mathbf{c} \times (\mathbf{a} \times \mathbf{b})$ on the right of $\eqref{eq:8-1}$. That is, in (7) replace $\mathbf{a} \to \mathbf{c}$, $\mathbf{b} \to \mathbf{a}$, $\mathbf{c} \to \mathbf{b}$.

\begin{equation}\mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{a}\,(\mathbf{c} \cdot \mathbf{b}) - \mathbf{b}\,(\mathbf{c} \cdot \mathbf{a}) \label{eq:8-2}\end{equation}

Substitute $\eqref{eq:8-2}$ into $\eqref{eq:8-1}$ and use commutativity of the dot product (1).

\begin{equation}(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = -[\mathbf{a}\,(\mathbf{b} \cdot \mathbf{c}) - \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c})] = \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c}) - \mathbf{a}\,(\mathbf{b} \cdot \mathbf{c}) \label{eq:8-3}\end{equation}

From $\eqref{eq:8-3}$, identity (8) is established.

Remark: Comparing (7) and (8) shows that the cross product is not associative: in general $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \ne (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$. The discrepancy is captured by the Jacobi identity (12).

1.4 Quadruple Products

This section treats products formed from four vectors, in particular the dot product of two cross products (Lagrange's identity) and the cross product of two cross products. They appear in spherical trigonometry, kinematics, and crystallography, among many other fields.

(9) Lagrange's identity

$$(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}) \tag{9}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \in \mathbb{R}^3$
Proof

Set $\mathbf{u} = \mathbf{a} \times \mathbf{b}$. Apply cyclic symmetry of the scalar triple product (5) to $\mathbf{u} \cdot (\mathbf{c} \times \mathbf{d})$.

\begin{equation}\mathbf{u} \cdot (\mathbf{c} \times \mathbf{d}) = \mathbf{d} \cdot (\mathbf{u} \times \mathbf{c}) \label{eq:9-1}\end{equation}

Restoring $\mathbf{u} = \mathbf{a} \times \mathbf{b}$,

\begin{equation}\mathbf{d} \cdot (\mathbf{u} \times \mathbf{c}) = \mathbf{d} \cdot ((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}) \label{eq:9-2}\end{equation}

Apply (8) to the right-hand side of $\eqref{eq:9-2}$.

\begin{equation}(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c}) - \mathbf{a}\,(\mathbf{b} \cdot \mathbf{c}) \label{eq:9-3}\end{equation}

Take the dot product of $\eqref{eq:9-3}$ with $\mathbf{d}$. Linearity allows the scalar coefficients to be pulled out.

\begin{equation}\mathbf{d} \cdot [\mathbf{b}\,(\mathbf{a} \cdot \mathbf{c}) - \mathbf{a}\,(\mathbf{b} \cdot \mathbf{c})] = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{b} \cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{d}) \label{eq:9-4}\end{equation}

Combine $\eqref{eq:9-1}$, $\eqref{eq:9-2}$, and $\eqref{eq:9-4}$.

\begin{equation}(\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}) \label{eq:9-5}\end{equation}

From $\eqref{eq:9-5}$, identity (9) holds.

Remark: Identity (9) can also be derived directly from the $\varepsilon$-$\delta$ identity. Writing $[(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d})] = \displaystyle\sum_{i,j,k,l,m} \varepsilon_{ijk} a_j b_k\, \varepsilon_{ilm} c_l d_m$ and applying $\varepsilon_{ijk}\varepsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$ gives the same result.

(10) Specialisation of Lagrange's identity

$$|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2\,|\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2 \tag{10}$$
Conditions: $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$
Proof

In Lagrange's identity (9), set $\mathbf{c} = \mathbf{a}$ and $\mathbf{d} = \mathbf{b}$.

\begin{equation}(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \cdot \mathbf{a})(\mathbf{b} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{b})(\mathbf{b} \cdot \mathbf{a}) \label{eq:10-1}\end{equation}

The left-hand side equals $|\mathbf{a} \times \mathbf{b}|^2$. On the right, $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$, $\mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2$, and by commutativity of the dot product (1) $\mathbf{b} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{b}$.

\begin{equation}|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2\,|\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2 \label{eq:10-2}\end{equation}

From $\eqref{eq:10-2}$, identity (10) is established.

Dot product a·b = |b|cosθ ×|a|; cross-product magnitude |a×b| = area of the parallelogram spanned by a,b; a×b is perpendicular to the plane Area = |a×b| = |a||b|sinθ proj |b|cosθ a b a × b (⊥ plane, right-handed) θ O Dot: a·b = |a| × (proj |b|cosθ) Cross: |a×b| = parallelogram area
Figure 1. Geometric meaning of the dot and cross products. The dot product $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ is the length of the projection of $\mathbf{b}$ onto $\mathbf{a}$ times $|\mathbf{a}|$, while the cross-product magnitude $|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta$ is the area of the parallelogram spanned by $\mathbf{a},\mathbf{b}$. The vector $\mathbf{a}\times\mathbf{b}$ is perpendicular to this plane, with the right-handed orientation.
Remark: Substituting $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}| \cos\theta$ into $\eqref{eq:10-2}$ gives $|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 (1 - \cos^2\theta) = |\mathbf{a}|^2 |\mathbf{b}|^2 \sin^2\theta$, hence $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\,\sin\theta$ ($0 \le \theta \le \pi$). This is the algebraic foundation of the geometric interpretation of the cross product as the area of the parallelogram.

(11) Cross product of two cross products

$$(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = \mathbf{c}\,(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{d})) - \mathbf{d}\,(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})) \tag{11}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \in \mathbb{R}^3$
Proof

Set $\mathbf{u} = \mathbf{a} \times \mathbf{b}$ and apply the BAC-CAB rule (7) to $\mathbf{u} \times (\mathbf{c} \times \mathbf{d})$. In (7) substitute $\mathbf{a} \to \mathbf{u}$, $\mathbf{b} \to \mathbf{c}$, $\mathbf{c} \to \mathbf{d}$.

\begin{equation}\mathbf{u} \times (\mathbf{c} \times \mathbf{d}) = \mathbf{c}\,(\mathbf{u} \cdot \mathbf{d}) - \mathbf{d}\,(\mathbf{u} \cdot \mathbf{c}) \label{eq:11-1}\end{equation}

Restore $\mathbf{u} = \mathbf{a} \times \mathbf{b}$.

\begin{equation}(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = \mathbf{c}\,((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}) - \mathbf{d}\,((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}) \label{eq:11-2}\end{equation}

By cyclic symmetry of the scalar triple product (5), $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d} = \mathbf{d} \cdot (\mathbf{a} \times \mathbf{b}) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{d})$, and similarly $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$.

\begin{equation}(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = \mathbf{c}\,(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{d})) - \mathbf{d}\,(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})) \label{eq:11-3}\end{equation}

From $\eqref{eq:11-3}$, identity (11) is established.

Remark: An alternative form is obtained by setting $\mathbf{v} = \mathbf{c} \times \mathbf{d}$ and applying (8) to $(\mathbf{a} \times \mathbf{b}) \times \mathbf{v}$, which gives $(\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = \mathbf{b}\,(\mathbf{a} \cdot (\mathbf{c} \times \mathbf{d})) - \mathbf{a}\,(\mathbf{b} \cdot (\mathbf{c} \times \mathbf{d}))$. The consistency of the two forms is related to the Jacobi identity (12).

1.5 Jacobi Identity

This section proves the Jacobi identity for the cross product. It is the essential relation that turns $\mathbb{R}^3$ into a Lie algebra under the cross product as the Lie bracket.

(12) Jacobi identity

$$\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \times (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{0} \tag{12}$$
Conditions: $\mathbf{a}, \mathbf{b}, \mathbf{c} \in \mathbb{R}^3$
Proof

Apply the BAC-CAB rule (7) to each of the three terms.

\begin{equation}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c}) - \mathbf{c}\,(\mathbf{a} \cdot \mathbf{b}) \label{eq:12-1}\end{equation}

For the second term, substitute $\mathbf{a} \to \mathbf{b}$, $\mathbf{b} \to \mathbf{c}$, $\mathbf{c} \to \mathbf{a}$.

\begin{equation}\mathbf{b} \times (\mathbf{c} \times \mathbf{a}) = \mathbf{c}\,(\mathbf{b} \cdot \mathbf{a}) - \mathbf{a}\,(\mathbf{b} \cdot \mathbf{c}) \label{eq:12-2}\end{equation}

For the third term, substitute $\mathbf{a} \to \mathbf{c}$, $\mathbf{b} \to \mathbf{a}$, $\mathbf{c} \to \mathbf{b}$.

\begin{equation}\mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{a}\,(\mathbf{c} \cdot \mathbf{b}) - \mathbf{b}\,(\mathbf{c} \cdot \mathbf{a}) \label{eq:12-3}\end{equation}

Sum $\eqref{eq:12-1}$, $\eqref{eq:12-2}$, and $\eqref{eq:12-3}$. Commutativity of the dot product (1) gives $\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$, $\mathbf{b}\cdot\mathbf{c} = \mathbf{c}\cdot\mathbf{b}$, and $\mathbf{c}\cdot\mathbf{a} = \mathbf{a}\cdot\mathbf{c}$.

\begin{equation}\begin{aligned}&\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \times (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) \\ &\quad= \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c}) - \mathbf{c}\,(\mathbf{a} \cdot \mathbf{b}) + \mathbf{c}\,(\mathbf{a} \cdot \mathbf{b}) - \mathbf{a}\,(\mathbf{b} \cdot \mathbf{c}) + \mathbf{a}\,(\mathbf{b} \cdot \mathbf{c}) - \mathbf{b}\,(\mathbf{a} \cdot \mathbf{c})\end{aligned} \label{eq:12-4}\end{equation}

On the right of $\eqref{eq:12-4}$, the terms with coefficients $\mathbf{a}, \mathbf{b}, \mathbf{c}$ cancel pairwise.

\begin{equation}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \times (\mathbf{c} \times \mathbf{a}) + \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{0} \label{eq:12-5}\end{equation}

From $\eqref{eq:12-5}$, identity (12) is established.

Remark: Identity (12) shows that $\mathbb{R}^3$ with the cross product $[\mathbf{a},\mathbf{b}] := \mathbf{a} \times \mathbf{b}$ as the Lie bracket forms the three-dimensional Lie algebra $\mathfrak{so}(3)$ (which is isomorphic to $\mathfrak{su}(2)$). The definition of a Lie algebra consists solely of antisymmetry (2) and the Jacobi identity (12).

References

Frequently Asked Questions

What are the main algebraic identities for the dot product?

Key identities include: commutativity $\mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}$, distributivity $\mathbf{a}\cdot(\mathbf{b}+\mathbf{c})=\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c}$, the Cauchy-Schwarz inequality $|\mathbf{a}\cdot\mathbf{b}|\leq|\mathbf{a}||\mathbf{b}|$, and the Lagrange identity $(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{c}\times\mathbf{d})=(\mathbf{a}\cdot\mathbf{c})(\mathbf{b}\cdot\mathbf{d})-(\mathbf{a}\cdot\mathbf{d})(\mathbf{b}\cdot\mathbf{c})$.

What are the main algebraic identities for the cross product?

Key identities: anti-commutativity $\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}$, the BAC-CAB rule $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{a}\cdot\mathbf{b})$, and the scalar triple product $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\det[\mathbf{a},\mathbf{b},\mathbf{c}]$.

What is the Levi-Civita symbol and how is it used in proofs?

The Levi-Civita symbol $\varepsilon_{ijk}$ equals $+1$ for even permutations of $(1,2,3)$, $-1$ for odd permutations, and $0$ when any two indices are equal. Writing $\mathbf{a}\times\mathbf{b}=\varepsilon_{ijk}a_j b_k\mathbf{e}_i$ transforms cross-product identities into index algebra, making complex vector identities straightforward to prove.