# Triple product

In geometry and algebra, the triple product is a product of three 3-dimensional vectors, usually Euclidean 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

The scalar triple product (also called the mixed product, box product, or triple scalar 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 unchanged under a circular shift of its three operands (a, b, c):
    $\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=\mathbf {b} \cdot (\mathbf {c} \times \mathbf {a} )=\mathbf {c} \cdot (\mathbf {a} \times \mathbf {b} )$
* Swapping the positions of the operators without re-ordering the operands leaves the triple product unchanged. This follows from the preceding property and the commutative property of the dot product:
    $\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c}$
* Swapping any two of the three operands negates the triple product. This follows from the circular-shift property and the anticommutativity of the cross product:
    ${\begin{aligned}\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )&=-\mathbf {a} \cdot (\mathbf {c} \times \mathbf {b} )\\&=-\mathbf {b} \cdot (\mathbf {a} \times \mathbf {c} )\\&=-\mathbf {c} \cdot (\mathbf {b} \times \mathbf {a} )\end{aligned}}$
* The scalar triple product can also be understood as the determinant of the 3×3 matrix that has the three vectors either as its rows or its columns (a matrix has the same determinant as its transpose):
    $\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=\det {\begin{bmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\\\end{bmatrix}}=\det {\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}=\det {\begin{bmatrix}\mathbf {a} &\mathbf {b} &\mathbf {c} \end{bmatrix}}.$
* 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 in the scalar triple product are equal, then its value is zero:
    $\mathbf {a} \cdot (\mathbf {a} \times \mathbf {b} )=\mathbf {a} \cdot (\mathbf {b} \times \mathbf {a} )=\mathbf {b} \cdot (\mathbf {a} \times \mathbf {a} )=0$
* Also:
    $(\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} ))\,\mathbf {a} =(\mathbf {a} \times \mathbf {b} )\times (\mathbf {a} \times \mathbf {c} )$
* The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products:$((\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} )\;((\mathbf {d} \times \mathbf {e} )\cdot \mathbf {f} )=\det {\begin{bmatrix}\mathbf {a} \cdot \mathbf {d} &\mathbf {a} \cdot \mathbf {e} &\mathbf {a} \cdot \mathbf {f} \\\mathbf {b} \cdot \mathbf {d} &\mathbf {b} \cdot \mathbf {e} &\mathbf {b} \cdot \mathbf {f} \\\mathbf {c} \cdot \mathbf {d} &\mathbf {c} \cdot \mathbf {e} &\mathbf {c} \cdot \mathbf {f} \end{bmatrix}}$This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. As a special case, the square of a triple product is a Gram determinant. Note that this determinant is well defined even when the vectors are in R$^m$ for m ≠ 3; in particular, the absolute value of a triple product can be computed for three vectors in R$^m$ by taking the square root of this determinant.
* The ratio of the triple product and the product of the three vector norms is known as a polar sine:${\frac {\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )}{\|{\mathbf {a} }\|\|{\mathbf {b} }\|\|{\mathbf {c} }\|}}=\operatorname {psin} (\mathbf {a} ,\mathbf {b} ,\mathbf {c} )$which ranges between −1 and 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 (of vectors) must be pseudoscalar-valued.
If T is a proper rotation then
    $\mathbf {Ta} \cdot (\mathbf {Tb} \times \mathbf {Tc} )=\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} ),$
but if T is an improper rotation then
    $\mathbf {Ta} \cdot (\mathbf {Tb} \times \mathbf {Tc} )=-\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} ).$

### Scalar or scalar density

Strictly speaking, a scalar does not change at all under a coordinate transformation. (For example, the factor of 2 used for doubling a vector does not change if the vector is in spherical vs. rectangular coordinates.) However, if each vector is transformed by a matrix then the triple product ends up being multiplied by the determinant of the transformation matrix, which could be quite arbitrary for a non-rotation. That is, the triple product is more properly described as a scalar density.

### As an exterior product

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, i.e.
    $|\mathbf {a} \wedge \mathbf {b} \wedge \mathbf {c} |=|\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )|$,
and is the Hodge dual of the scalar triple product. As the exterior product is associative brackets are not needed as it does not matter which of a ∧ b or b ∧ c is calculated first, though the order of the vectors in the product does matter. Geometrically the trivector a ∧ b ∧ c corresponds to the parallelepiped spanned by a, b, and c, with bivectors a ∧ b, b ∧ c and a ∧ c matching the parallelogram faces of the parallelepiped.

### As a trilinear function

The triple product is identical to the volume form of the Euclidean 3-space applied to the vectors via interior product. It also can be expressed as a contraction of vectors with a rank-3 tensor equivalent to the form (or a pseudotensor equivalent to the volume pseudoform); see below.

## 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:
    $\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c}$.
This is known as triple product expansion, or Lagrange's formula, although the latter name is also used for several other formulas. Its right hand side can be remembered by using the mnemonic "ACB − ABC", provided one keeps in mind which vectors are dotted together. A proof is provided below. Some textbooks write the identity as $\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=\mathbf {b} (\mathbf {a} \cdot \mathbf {c} )-\mathbf {c} (\mathbf {a} \cdot \mathbf {b} )$ such that a more familiar mnemonic "BAC − CAB" is obtained, as in “back of the cab”.
Since the cross product is anticommutative, this formula may also be written (up to permutation of the letters) as:
    $(\mathbf {a} \times \mathbf {b} )\times \mathbf {c} =-\mathbf {c} \times (\mathbf {a} \times \mathbf {b} )=-(\mathbf {c} \cdot \mathbf {b} )\mathbf {a} +(\mathbf {c} \cdot \mathbf {a} )\mathbf {b}$
From Lagrange's formula it follows that the vector triple product satisfies:
    $\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )+\mathbf {b} \times (\mathbf {c} \times \mathbf {a} )+\mathbf {c} \times (\mathbf {a} \times \mathbf {b} )=\mathbf {0}$
which is the Jacobi identity for the cross product. Another useful formula follows:
    $(\mathbf {a} \times \mathbf {b} )\times \mathbf {c} =\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )-\mathbf {b} \times (\mathbf {a} \times \mathbf {c} )$
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:
    ${\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\times \mathbf {A} )={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\cdot \mathbf {A} )-({\boldsymbol {\nabla }}\cdot {\boldsymbol {\nabla }})\mathbf {A}$
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:
    ${\begin{aligned}(\mathbf {u} \times (\mathbf {v} \times \mathbf {w} ))_{x}&=\mathbf {u} _{y}(\mathbf {v} _{x}\mathbf {w} _{y}-\mathbf {v} _{y}\mathbf {w} _{x})-\mathbf {u} _{z}(\mathbf {v} _{z}\mathbf {w} _{x}-\mathbf {v} _{x}\mathbf {w} _{z})\\&=\mathbf {v} _{x}(\mathbf {u} _{y}\mathbf {w} _{y}+\mathbf {u} _{z}\mathbf {w} _{z})-\mathbf {w} _{x}(\mathbf {u} _{y}\mathbf {v} _{y}+\mathbf {u} _{z}\mathbf {v} _{z})\\&=\mathbf {v} _{x}(\mathbf {u} _{y}\mathbf {w} _{y}+\mathbf {u} _{z}\mathbf {w} _{z})-\mathbf {w} _{x}(\mathbf {u} _{y}\mathbf {v} _{y}+\mathbf {u} _{z}\mathbf {v} _{z})+(\mathbf {u} _{x}\mathbf {v} _{x}\mathbf {w} _{x}-\mathbf {u} _{x}\mathbf {v} _{x}\mathbf {w} _{x})\\&=\mathbf {v} _{x}(\mathbf {u} _{x}\mathbf {w} _{x}+\mathbf {u} _{y}\mathbf {w} _{y}+\mathbf {u} _{z}\mathbf {w} _{z})-\mathbf {w} _{x}(\mathbf {u} _{x}\mathbf {v} _{x}+\mathbf {u} _{y}\mathbf {v} _{y}+\mathbf {u} _{z}\mathbf {v} _{z})\\&=(\mathbf {u} \cdot \mathbf {w} )\mathbf {v} _{x}-(\mathbf {u} \cdot \mathbf {v} )\mathbf {w} _{x}\end{aligned}}$
Similarly, the $y$ and $z$ components of $\mathbf {u} \times (\mathbf {v} \times \mathbf {w} )$ are given by:
    ${\begin{aligned}(\mathbf {u} \times (\mathbf {v} \times \mathbf {w} ))_{y}&=(\mathbf {u} \cdot \mathbf {w} )\mathbf {v} _{y}-(\mathbf {u} \cdot \mathbf {v} )\mathbf {w} _{y}\\(\mathbf {u} \times (\mathbf {v} \times \mathbf {w} ))_{z}&=(\mathbf {u} \cdot \mathbf {w} )\mathbf {v} _{z}-(\mathbf {u} \cdot \mathbf {v} )\mathbf {w} _{z}\end{aligned}}$
By combining these three components we obtain:
    $\mathbf {u} \times (\mathbf {v} \times \mathbf {w} )=(\mathbf {u} \cdot \mathbf {w} )\ \mathbf {v} -(\mathbf {u} \cdot \mathbf {v} )\ \mathbf {w}$

### Using geometric algebra

If geometric algebra is used the cross product b × c of vectors is expressed as their exterior product b∧c, a bivector. The second cross product cannot be expressed as an exterior product, otherwise the scalar triple product would result. Instead a left contraction can be used, so the formula becomes
    ${\begin{aligned}-\mathbf {a} \;{\big \lrcorner }\;(\mathbf {b} \wedge \mathbf {c} )&=\mathbf {b} \wedge (\mathbf {a} \;{\big \lrcorner }\;\mathbf {c} )-(\mathbf {a} \;{\big \lrcorner }\;\mathbf {b} )\wedge \mathbf {c} \\&=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} \end{aligned}}$
The proof follows from the properties of the contraction. The result is the same vector as calculated using a × (b × c).

## Triple Bivector Product

In geometric algebra, three bivectors can also have a triple product. This product mimic the standard triple vector product. The antisymmetric product of three bivectors is.
    ${\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\times {\overset {\Rightarrow }{c}})=-({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{c}}){\overset {\Rightarrow }{b}}+({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{b}}){\overset {\Rightarrow }{c}}$

### Proof

This proof is made by taking dual of the geometric algebra version of the triple vector product until all vectors become bivectors.
    ${\begin{aligned}(-\mathbf {a} \;{\big \lrcorner }\;(\mathbf {b} \wedge \mathbf {c} ))\star &=-{\frac {1}{2}}({\overset {\Rightarrow }{a}}(\mathbf {b} \wedge \mathbf {c} )-(\mathbf {b} \wedge \mathbf {c} ){\overset {\Rightarrow }{a}})=-{\overset {\Rightarrow }{a}}\times (\mathbf {b} \wedge \mathbf {c} )\\(-{\overset {\Rightarrow }{a}}\times (\mathbf {b} \wedge \mathbf {c} ))\star &=-{\frac {1}{2}}({\overset {\Rightarrow }{a}}{\frac {1}{2}}({\overset {\Rightarrow }{b}}\mathbf {c} -\mathbf {c} {\overset {\Rightarrow }{b}})-{\frac {1}{2}}({\overset {\Rightarrow }{b}}\mathbf {c} -\mathbf {c} {\overset {\Rightarrow }{b}}){\overset {\Rightarrow }{a}})=-{\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\cdot \mathbf {c} )\\(-{\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\cdot \mathbf {c} ))\star &=-{\frac {1}{2}}({\overset {\Rightarrow }{a}}{\frac {1}{2}}({\overset {\Rightarrow }{b}}{\overset {\Rightarrow }{c}}-{\overset {\Rightarrow }{c}}{\overset {\Rightarrow }{b}})-{\frac {1}{2}}({\overset {\Rightarrow }{b}}{\overset {\Rightarrow }{c}}-{\overset {\Rightarrow }{c}}{\overset {\Rightarrow }{b}}){\overset {\Rightarrow }{a}})={\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\times {\overset {\Rightarrow }{c}})\end{aligned}}$
This was three duals. This must also be done to the left side.
    ${\begin{aligned}&((((\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} )\star )\star )\star =\\&({\frac {1}{2}}({\overset {\Rightarrow }{a}}{\overset {\Rightarrow }{c}}+{\overset {\Rightarrow }{c}}{\overset {\Rightarrow }{a}})){\overset {\Rightarrow }{b}}-({\frac {1}{2}}({\overset {\Rightarrow }{a}}{\overset {\Rightarrow }{b}}+{\overset {\Rightarrow }{b}}{\overset {\Rightarrow }{a}})){\overset {\Rightarrow }{c}}=\\&({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{c}}){\overset {\Rightarrow }{b}}+({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{b}}){\overset {\Rightarrow }{c}}\end{aligned}}$
By negating both side we obtain:
    ${\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\times {\overset {\Rightarrow }{c}})=-({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{c}}){\overset {\Rightarrow }{b}}+({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{b}}){\overset {\Rightarrow }{c}}$

## Interpretations
### Tensor calculus

In tensor notation, the triple product is expressed using the Levi-Civita symbol: $\mathbf {a} \cdot [\mathbf {b} \times \mathbf {c} ]=\varepsilon _{ijk}a^{i}b^{j}c^{k}$ and $(\mathbf {a} \times [\mathbf {b} \times \mathbf {c} ])_{i}=\varepsilon _{ijk}a^{j}\varepsilon ^{k\ell m}b_{\ell }c_{m}=\varepsilon _{ijk}\varepsilon ^{k\ell m}a^{j}b_{\ell }c_{m},$ referring to the $i$-th component of the resulting vector. This can be simplified by performing a contraction on the Levi-Civita symbols, $\varepsilon _{ijk}\varepsilon ^{k\ell m}=\delta _{ij}^{\ell m}=\delta _{i}^{\ell }\delta _{j}^{m}-\delta _{i}^{m}\delta _{j}^{\ell }\,,$ where $\delta _{j}^{i}$ is the Kronecker delta function ($\delta _{j}^{i}=0$ when $i\neq j$ and $\delta _{j}^{i}=1$ when $i=j$) and $\delta _{ij}^{\ell m}$ is the generalized Kronecker delta function. We can reason out this identity by recognizing that the index $k$ will be summed out leaving only $i$ and $j$. In the first term, we fix $i=l$ and thus $j=m$. Likewise, in the second term, we fix $i=m$ and thus $l=j$.
Returning to the triple cross product, $(\mathbf {a} \times [\mathbf {b} \times \mathbf {c} ])_{i}=(\delta _{i}^{\ell }\delta _{j}^{m}-\delta _{i}^{m}\delta _{j}^{\ell })a^{j}b_{\ell }c_{m}=a^{j}b_{i}c_{j}-a^{j}b_{j}c_{i}=b_{i}(\mathbf {a} \cdot \mathbf {c} )-c_{i}(\mathbf {a} \cdot \mathbf {b} )\,.$

### Vector calculus

Consider the flux integral of the vector field $\mathbf {F}$ across the parametrically-defined surface $S=\mathbf {r} (u,v)$: ∬$_S$F⋅n$^^$dS{\textstyle \iint _{S}\mathbf {F} \cdot {\hat {\mathbf {n} }}\, dS}. The unit normal vector ${\hat {\mathbf {n} }}$ to the surface is given by r$_u$×r$_v$/|r$_u$×r$_v$| {\textstyle {\frac {\mathbf {r} _{u}\times \mathbf {r} _{v}}{|\mathbf {r} _{u}\times \mathbf {r} _{v}|}}}, so the integrand F⋅ (r$_u$×r$_v$)/|r$_u$×r$_v$| {\textstyle \mathbf {F} \cdot {\frac {(\mathbf {r} _{u}\times \mathbf {r} _{v})}{|\mathbf {r} _{u}\times \mathbf {r} _{v}|}}} is a scalar triple product.