# Triple product

(Redirected from Signed volume)

In mathematics, the triple product is a product of three vectors. The name "triple product" is used for two different products, the scalar-valued scalar triple product and, less often, the vector-valued vector triple product.

## Scalar triple product

Three vectors defining a parallelepiped

The scalar triple product (also called the mixed or box product) is defined as the dot product of one of the vectors with the cross product of the other two.

### Geometric interpretation

Geometrically, the scalar triple product

$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})$

is the (signed) volume of the parallelepiped defined by the three vectors given.

### Properties

• The scalar triple product is invariant under a circular shift of its three operands (a, b, c):
$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})= \mathbf{b}\cdot(\mathbf{c}\times \mathbf{a})= \mathbf{c}\cdot(\mathbf{a}\times \mathbf{b})$
• Switching the two vectors in the cross product negates the triple product:
$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c}) = -\mathbf{a}\cdot(\mathbf{c}\times \mathbf{b})$
Here, the parentheses may be omitted without causing ambiguity, since the dot product cannot be evaluated first. If it were, it would leave the cross product of a scalar and a vector, which is not defined.
• The scalar triple product can also be understood as the determinant of the 3 × 3 matrix having the three vectors either as its rows or its columns (a matrix has the same determinant as its transpose):
$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c}) = \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{bmatrix}.$
$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{Ra} \cdot (\mathbf{Rb} \times \mathbf{Rc})$
• If the scalar triple product is equal to zero, then the three vectors a, b, and c are coplanar, since the "parallelepiped" defined by them would be flat and have no volume.
• If any two vectors of triple scalar product are equal, then its value is zero:
$\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{a}) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{b}) = \mathbf{a} \cdot (\mathbf{a} \times \mathbf{a}) = 0$
• Moreover,
$[\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})] \mathbf{a} = (\mathbf{a}\times \mathbf{b})\times (\mathbf{a}\times \mathbf{c})$
• The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products
$((\mathbf{a}\times \mathbf{b})\cdot \mathbf{c})\;((\mathbf{d}\times \mathbf{e})\cdot \mathbf{f}) = \det\left[ \begin{pmatrix} \mathbf{a} \\ \mathbf{b} \\ \mathbf{c} \end{pmatrix}\cdot \begin{pmatrix} \mathbf{d} & \mathbf{e} & \mathbf{f} \end{pmatrix}\right] = \det \begin{bmatrix} \mathbf{a}\cdot \mathbf{d} & \mathbf{a}\cdot \mathbf{e} & \mathbf{a}\cdot \mathbf{f} \\ \mathbf{b}\cdot \mathbf{d} & \mathbf{b}\cdot \mathbf{e} & \mathbf{b}\cdot \mathbf{f} \\ \mathbf{c}\cdot \mathbf{d} & \mathbf{c}\cdot \mathbf{e} & \mathbf{c}\cdot \mathbf{f} \end{bmatrix}$[1]

### Scalar or pseudoscalar

Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the parity of the permutation of the vectors. This means the product is negated if the orientation is reversed, for example by a parity transformation, and so is more properly described as a pseudoscalar if the orientation can change.

This also relates to the handedness of the cross product; the cross product transforms as a pseudovector under parity transformations and so is properly described as a pseudovector. The dot product of two vectors is a scalar but the dot product of a pseudovector and a vector is a pseudoscalar, so the scalar triple product must be pseudoscalar valued.

### As an exterior product

A trivector is an oriented volume element; its Hodge dual is a scalar with magnitude equal to its volume.

In exterior algebra and geometric algebra the exterior product of two vectors is a bivector, while the exterior product of three vectors is a trivector. A bivector is an oriented plane element and a trivector is an oriented volume element, in the same way that a vector is an oriented line element. Given vectors a, b and c, the product

$\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}$

is a trivector with magnitude equal to the scalar triple product, and is the pseudoscalar dual of the triple product. As the exterior product is associative brackets are not needed as it does not matter which of ab or bc is calculated first, though the order of the vectors in the product does matter. Geometrically the trivector abc corresponds to the parallelepiped spanned by a, b, and c, with bivectors ab, bc and ac matching the parallelogram faces of the parallelepiped.

## Vector triple product

The vector triple product is defined as the cross product of one vector with the cross product of the other two. The following relationship holds:

$\mathbf{a}\times (\mathbf{b}\times \mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})$.

This known as triple product expansion, or Lagrange's formula,[2][3] although the latter name is also used for several other formulae. Its right hand side can be remembered by using the mnemonic "BAC–CAB", provided one keeps in mind which vectors are dotted together. A proof is provided below.

Since the cross product is anticommutative, the following related formula can be easily derived:

$(\mathbf{a}\times \mathbf{b})\times \mathbf{c} = -\mathbf{c}\times(\mathbf{a}\times \mathbf{b}) = -(\mathbf{c}\cdot\mathbf{b})\mathbf{a} + (\mathbf{c}\cdot\mathbf{a})\mathbf{b}$

The vector triple product also satisfies the Jacobi identity:

$\mathbf{a}\times (\mathbf{b}\times \mathbf{c}) \; + \mathbf{b}\times (\mathbf{c}\times \mathbf{a}) \; + \mathbf{c}\times (\mathbf{a}\times \mathbf{b}) = 0$

These formulas are very useful in simplifying vector calculations in physics. A related identity regarding gradients and useful in vector calculus is Lagrange's formula of vector cross-product identity:[4]

\begin{align} \nabla \times (\nabla \times \mathbf{f}) & {}= \nabla (\nabla \cdot \mathbf{f} ) - (\nabla \cdot \nabla) \mathbf{f} \\ & {}= \mbox{grad }(\mbox{div } \mathbf{f} ) - \nabla^2 \mathbf{f}. \end{align}

This can be also regarded as a special case of the more general Laplace-de Rham operator $\Delta = d \delta + \delta d$.

### Proof

The $x$ component of $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$ is given by:

$\mathbf{u}_y(\mathbf{v}_x\mathbf{w}_y-\mathbf{v}_y\mathbf{w}_x)-\mathbf{u}_z(\mathbf{v}_z\mathbf{w}_x-\mathbf{v}_x\mathbf{w}_z)$

or

$\mathbf{v}_x(\mathbf{u}_y\mathbf{w}_y+\mathbf{u}_z\mathbf{w}_z)-\mathbf{w}_x(\mathbf{u}_y\mathbf{v}_y+\mathbf{u}_z\mathbf{v}_z)$

By adding and subtracting $\mathbf{u}_x\mathbf{v}_x\mathbf{w}_x$, this becomes

$\mathbf{v}_x(\mathbf{u}_x\mathbf{w}_x+\mathbf{u}_y\mathbf{w}_y+\mathbf{u}_z\mathbf{w}_z)-\mathbf{w}_x(\mathbf{u}_x\mathbf{v}_x+\mathbf{u}_y\mathbf{v}_y+\mathbf{u}_z\mathbf{v}_z)=(\mathbf{u}\cdot\mathbf{w})\mathbf{v}_x-(\mathbf{u}\cdot\mathbf{v})\mathbf{w}_x$

Similarly, the $y$ and $z$ components of $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$ are given by:

$(\mathbf{u}\cdot\mathbf{w})\mathbf{v}_y-(\mathbf{u}\cdot\mathbf{v})\mathbf{w}_y$

and

$(\mathbf{u}\cdot\mathbf{w})\mathbf{v}_z-(\mathbf{u}\cdot\mathbf{v})\mathbf{w}_z$

By combining these three components we obtain:

$\mathbf{u}\times (\mathbf{v}\times \mathbf{w}) = (\mathbf{u}\cdot\mathbf{w})\ \mathbf{v} - (\mathbf{u}\cdot\mathbf{v})\ \mathbf{w}$[5]

### Vector or pseudovector

Where parity transformations need to be considered, so the cross product is treated as a pseudovector, the vector triple product is vector rather than pseudovector valued, as it is the product of a vector a and a pseudovector b × c. This can also be seen from the expansion in terms of the dot product, which consists only of a sum of vectors multiplied by scalars so must be vector valued.

## Notation

Using the Levi-Civita symbol, the triple product is

$(\mathbf{a} \cdot (\mathbf{b}\times \mathbf{c})) = \varepsilon_{ijk} a^i b^j c^k$

and

$(\mathbf{a} \times (\mathbf{b}\times \mathbf{c}))_i = \varepsilon_{ijk} a^j \varepsilon_{k\ell m} b^\ell c^m = \varepsilon_{ijk}\varepsilon_{k\ell m} a^j b^\ell c^m$

which can be simplified by performing a contraction on the Levi-Civita symbols, $\varepsilon_{ijk} \varepsilon_{k\ell m}=\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{\ell j}$ and simplifying the result.