Triple product

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:

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

This is known as triple product expansion, or Lagrange's formula, although the latter name is also used for several other formulas. A proof is provided below.

Since the cross product is anticommutative, this formula may also be written (up to permutation of the letters) as:

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

From Lagrange's formula it follows that the vector triple product satisfies:

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

which is the Jacobi identity for the cross product. Another useful formula follows:

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

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:

${\displaystyle \mathbf{\nabla }\times (\mathbf{\nabla }\times \mathbf{A})=\mathbf{\nabla }(\mathbf{\nabla }\cdot \mathbf{A})-(\mathbf{\nabla }\cdot \mathbf{\nabla })\mathbf{A}}$

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

Proof

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

${\displaystyle \begin{array}{cc}(\mathbf{u}\times (\mathbf{v}\times \mathbf{w}){)}_x& ={\mathbf{u}}_y({\mathbf{v}}_x{\mathbf{w}}_y-{\mathbf{v}}_y{\mathbf{w}}_x)-{\mathbf{u}}_z({\mathbf{v}}_z{\mathbf{w}}_x-{\mathbf{v}}_x{\mathbf{w}}_z)\\ & ={\mathbf{v}}_x({\mathbf{u}}_y{\mathbf{w}}_y+{\mathbf{u}}_z{\mathbf{w}}_z)-{\mathbf{w}}_x({\mathbf{u}}_y{\mathbf{v}}_y+{\mathbf{u}}_z{\mathbf{v}}_z)\\ & ={\mathbf{v}}_x({\mathbf{u}}_y{\mathbf{w}}_y+{\mathbf{u}}_z{\mathbf{w}}_z)-{\mathbf{w}}_x({\mathbf{u}}_y{\mathbf{v}}_y+{\mathbf{u}}_z{\mathbf{v}}_z)+({\mathbf{u}}_x{\mathbf{v}}_x{\mathbf{w}}_x-{\mathbf{u}}_x{\mathbf{v}}_x{\mathbf{w}}_x)\\ & ={\mathbf{v}}_x({\mathbf{u}}_x{\mathbf{w}}_x+{\mathbf{u}}_y{\mathbf{w}}_y+{\mathbf{u}}_z{\mathbf{w}}_z)-{\mathbf{w}}_x({\mathbf{u}}_x{\mathbf{v}}_x+{\mathbf{u}}_y{\mathbf{v}}_y+{\mathbf{u}}_z{\mathbf{v}}_z)\\ & =(\mathbf{u}\cdot \mathbf{w}){\mathbf{v}}_x-(\mathbf{u}\cdot \mathbf{v}){\mathbf{w}}_x\end{array}}$

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

${\displaystyle \begin{array}{cc}(\mathbf{u}\times (\mathbf{v}\times \mathbf{w}){)}_y& =(\mathbf{u}\cdot \mathbf{w}){\mathbf{v}}_y-(\mathbf{u}\cdot \mathbf{v}){\mathbf{w}}_y\\ (\mathbf{u}\times (\mathbf{v}\times \mathbf{w}){)}_z& =(\mathbf{u}\cdot \mathbf{w}){\mathbf{v}}_z-(\mathbf{u}\cdot \mathbf{v}){\mathbf{w}}_z\end{array}}$

By combining these three components we obtain:

${\displaystyle \mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}\cdot \mathbf{w})\mathbf{v}-(\mathbf{u}\cdot \mathbf{v})\mathbf{w}}$

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

${\displaystyle \begin{array}{cc}-\mathbf{a}⌟(\mathbf{b}\wedge \mathbf{c})& =\mathbf{b}\wedge (\mathbf{a}⌟\mathbf{c})-(\mathbf{a}⌟\mathbf{b})\wedge \mathbf{c}\\ & =(\mathbf{a}\cdot \mathbf{c})\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{c}\end{array}}$

The proof follows from the properties of the contraction. The result is the same vector as calculated using a × (b × c).

Non-commutative triple vector product

With non-commutative vector operators ${\displaystyle \hat{\mathbf{a}},\hat{\mathbf{b}},\hat{\mathbf{c}}}$, special relations hold for the triple product in accordance to

${\displaystyle \hat{\mathbf{a}}\times \left(\hat{\mathbf{b}}\times \hat{\mathbf{c}}\right)=\hat{\mathbf{b}}(\hat{\mathbf{a}}\cdot \hat{\mathbf{c}})-\hat{\mathbf{c}}(\hat{\mathbf{a}}\cdot \hat{\mathbf{b}})+\sum _{\alpha ,\beta =1}^3\left(\left[{\hat{a}}_{\beta },{\hat{b}}_{\alpha }\right]{\hat{c}}_{\beta }+\left[{\hat{c}}_{\alpha },{\hat{a}}_{\beta }\right]{\hat{b}}_{\beta }+{\hat{a}}_{\beta }\left[{\hat{c}}_{\alpha },{\hat{b}}_{\beta }\right]\right){\mathbf{e}}_{\alpha }}$${\displaystyle \left(\hat{\mathbf{a}}\times \hat{\mathbf{b}}\right)\times \hat{\mathbf{c}}=\hat{\mathbf{b}}\left(\hat{\mathbf{a}}\cdot \hat{\mathbf{c}}\right)-\hat{\mathbf{a}}\left(\hat{\mathbf{b}}\cdot \hat{\mathbf{c}}\right)+\sum _{\alpha ,\beta =1}^3\left[{\hat{a}}_{\beta },{\hat{b}}_{\alpha }\right]{\hat{c}}_{\beta }{\mathbf{e}}_{\alpha }}$

with the unit vectors ${\displaystyle {\mathbf{e}}_{\alpha }}$of ${\displaystyle \alpha }$, with ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ being indices of a three dimensional orthonormal basis and the square brackets repressenting the commutator.

Proof

Using the individual components of the cross product

${\displaystyle {\left(\mathbf{a}\times \mathbf{b}\right)}_{\alpha }={\epsilon }_{\alpha \beta \gamma }{a}_{\beta }{b}_{\gamma },}$

as well as the Levi-Civita symbol expression via the Kronecker Delta

${\displaystyle {\epsilon }_{\alpha \beta \gamma }{\epsilon }_{\gamma \mu \nu }={\delta }_{\alpha \mu }{\delta }_{\beta \nu }-{\delta }_{\alpha \nu }{\delta }_{\beta \mu },}$

we receive the expression

${\displaystyle \begin{array}{cc}{\left(\hat{\mathbf{a}}\times \left(\hat{\mathbf{b}}\times \hat{\mathbf{c}}\right)\right)}_{\alpha }& ={\epsilon }_{\alpha \beta \gamma }{\hat{a}}_{\beta }{\left(\hat{\mathbf{b}}\times \hat{\mathbf{c}}\right)}_{\gamma }={\epsilon }_{\alpha \beta \gamma }{\hat{a}}_{\beta }{\epsilon }_{\gamma \mu \nu }{\hat{b}}_{\mu }{\hat{c}}_{\nu }\\ & =\left({\delta }_{\alpha \mu }{\delta }_{\beta \nu }-{\delta }_{\alpha \nu }{\delta }_{\beta \mu }\right){\hat{a}}_{\beta }{\hat{b}}_{\mu }{\hat{c}}_{\nu }={\hat{a}}_{\beta }{\hat{b}}_{\alpha }{\hat{c}}_{\beta }-{\hat{a}}_{\beta }{\hat{b}}_{\beta }{\hat{c}}_{\alpha }\\ & =\left({\hat{b}}_{\alpha }{\hat{a}}_{\beta }{\hat{c}}_{\beta }-{\hat{c}}_{\alpha }{\hat{a}}_{\beta }{\hat{b}}_{\beta }\right)+\left({\hat{a}}_{\beta }{\hat{b}}_{\alpha }{\hat{c}}_{\beta }-{\hat{b}}_{\alpha }{\hat{a}}_{\beta }{\hat{c}}_{\beta }\right)+\left({\hat{c}}_{\alpha }{\hat{a}}_{\beta }{\hat{b}}_{\beta }-{\hat{a}}_{\beta }{\hat{b}}_{\beta }{\hat{c}}_{\alpha }\right)\\ & ={\hat{b}}_{\alpha }\left({\hat{a}}_{\beta }{\hat{c}}_{\beta }\right)-{\hat{c}}_{\alpha }\left({\hat{a}}_{\beta }{\hat{b}}_{\beta }\right)+\left[{\hat{a}}_{\beta },{\hat{b}}_{\alpha }\right]{\hat{c}}_{\beta }+\left[{\hat{c}}_{\alpha },{\hat{a}}_{\beta }{\hat{b}}_{\beta }\right]\\ & ={\hat{b}}_{\alpha }\left(\hat{\mathbf{a}}\cdot \hat{\mathbf{c}}\right)-{\hat{c}}_{\alpha }\left(\hat{\mathbf{a}}\cdot \hat{\mathbf{b}}\right)+\left[{\hat{a}}_{\beta },{\hat{b}}_{\alpha }\right]{\hat{c}}_{\beta }+\left[{\hat{c}}_{\alpha },{\hat{a}}_{\beta }\right]{\hat{b}}_{\beta }+a_{\beta }\left[{\hat{c}}_{\alpha },{\hat{b}}_{\beta }\right]\end{array}}$

for the first identity and the expression

${\displaystyle \begin{array}{cc}{\left(\left(\hat{\mathbf{a}}\times \hat{\mathbf{b}}\right)\times \hat{\mathbf{c}}\right)}_{\alpha }& ={\epsilon }_{\alpha \beta \gamma }{\left(\hat{\mathbf{a}}\times \hat{\mathbf{b}}\right)}_{\beta }{\hat{c}}_{\gamma }={\epsilon }_{\alpha \beta \gamma }{\epsilon }_{\beta \mu \nu }{\hat{a}}_{\mu }{\hat{b}}_{\nu }{\hat{c}}_{\gamma }\\ & =\left({\delta }_{\nu \alpha }{\delta }_{\mu \gamma }-{\delta }_{\mu \alpha }{\delta }_{\nu \gamma }\right){\hat{a}}_{\mu }{\hat{b}}_{\nu }{\hat{c}}_{\gamma }={\hat{a}}_{\mu }{\hat{b}}_{\alpha }{\hat{c}}_{\mu }-{\hat{a}}_{\alpha }{\hat{b}}_{\nu }{\hat{c}}_{\nu }\\ & =\left({\hat{a}}_{\mu }{\hat{b}}_{\alpha }-{\hat{b}}_{\alpha }{\hat{a}}_{\mu }+{\hat{b}}_{\alpha }{\hat{a}}_{\mu }\right){\hat{c}}_{\mu }-{\hat{a}}_{\alpha }\left(\hat{\mathbf{b}}\cdot \hat{\mathbf{c}}\right)\\ & =\left(\left[{\hat{a}}_{\mu },{\hat{b}}_{\alpha }\right]+{\hat{b}}_{\alpha }{\hat{a}}_{\mu }\right){\hat{c}}_{\mu }-{\hat{a}}_{\alpha }\left(\hat{\mathbf{b}}\cdot \hat{\mathbf{c}}\right)\\ & ={\hat{b}}_{\alpha }\left(\hat{\mathbf{a}}\cdot \hat{\mathbf{c}}\right)-{\hat{a}}_{\alpha }\left(\hat{\mathbf{b}}\cdot \hat{\mathbf{c}}\right)+\left[{\hat{a}}_{\mu },{\hat{b}}_{\alpha }\right]{\hat{c}}_{\mu }\end{array}}$

for the second identity for each of the three indices. By expressing them via a sum of all indices, the original identity is received.

Triple bivector product

In geometric algebra, three bivectors can also have a triple product. This product mimics the standard triple vector product. The antisymmetric product of three bivectors is.

${\displaystyle \stackrel{\Rightarrow }{a}\times \left(\stackrel{\Rightarrow }{b}\times \stackrel{\Rightarrow }{c}\right)=-\left(\stackrel{\Rightarrow }{a}\cdot \stackrel{\Rightarrow }{c}\right)\stackrel{\Rightarrow }{b}+\left(\stackrel{\Rightarrow }{a}\cdot \stackrel{\Rightarrow }{b}\right)\stackrel{\Rightarrow }{c}}$

Proof

This proof is made by taking dual of the geometric algebra version of the triple vector product until all vectors become bivectors.

${\displaystyle \begin{array}{rlrlrl}& (-\mathbf{a}⌟(\mathbf{b}\wedge \mathbf{c}))\star & & =-\frac{1}{2}\left(\stackrel{\Rightarrow }{a}(\mathbf{b}\wedge \mathbf{c})-(\mathbf{b}\wedge \mathbf{c})\stackrel{\Rightarrow }{a}\right)& & =-\stackrel{\Rightarrow }{a}\times (\mathbf{b}\wedge \mathbf{c})\\ & (-\stackrel{\Rightarrow }{a}\times (\mathbf{b}\wedge \mathbf{c}))\star & & =-\frac{1}{2}\left(\stackrel{\Rightarrow }{a}\frac{1}{2}(\stackrel{\Rightarrow }{b}\mathbf{c}-\mathbf{c}\stackrel{\Rightarrow }{b})-\frac{1}{2}(\stackrel{\Rightarrow }{b}\mathbf{c}-\mathbf{c}\stackrel{\Rightarrow }{b})\stackrel{\Rightarrow }{a}\right)& & =-\stackrel{\Rightarrow }{a}\times (\stackrel{\Rightarrow }{b}\cdot \mathbf{c})\\ & (-\stackrel{\Rightarrow }{a}\times (\stackrel{\Rightarrow }{b}\cdot \mathbf{c}))\star & & =-\frac{1}{2}\left(\stackrel{\Rightarrow }{a}\frac{1}{2}(\stackrel{\Rightarrow }{b}\stackrel{\Rightarrow }{c}-\stackrel{\Rightarrow }{c}\stackrel{\Rightarrow }{b})-\frac{1}{2}(\stackrel{\Rightarrow }{b}\stackrel{\Rightarrow }{c}-\stackrel{\Rightarrow }{c}\stackrel{\Rightarrow }{b})\stackrel{\Rightarrow }{a}\right)& & =\stackrel{\Rightarrow }{a}\times (\stackrel{\Rightarrow }{b}\times \stackrel{\Rightarrow }{c})\end{array}}$

This was three duals. This must also be done to the left side.

${\displaystyle \begin{array}{cc}& ((((\mathbf{a}\cdot \mathbf{c})\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{c})\star )\star )\star \\ & =\frac{1}{2}(\stackrel{\Rightarrow }{a}\stackrel{\Rightarrow }{c}+\stackrel{\Rightarrow }{c}\stackrel{\Rightarrow }{a})\stackrel{\Rightarrow }{b}-\frac{1}{2}(\stackrel{\Rightarrow }{a}\stackrel{\Rightarrow }{b}+\stackrel{\Rightarrow }{b}\stackrel{\Rightarrow }{a})\stackrel{\Rightarrow }{c}\\ & =(\stackrel{\Rightarrow }{a}\cdot \stackrel{\Rightarrow }{c})\stackrel{\Rightarrow }{b}+(\stackrel{\Rightarrow }{a}\cdot \stackrel{\Rightarrow }{b})\stackrel{\Rightarrow }{c}\end{array}}$

By negating both side we obtain:

${\displaystyle \stackrel{\Rightarrow }{a}\times (\stackrel{\Rightarrow }{b}\times \stackrel{\Rightarrow }{c})=-(\stackrel{\Rightarrow }{a}\cdot \stackrel{\Rightarrow }{c})\stackrel{\Rightarrow }{b}+(\stackrel{\Rightarrow }{a}\cdot \stackrel{\Rightarrow }{b})\stackrel{\Rightarrow }{c}}$

Triple products using tensor notation

It can be useful in fields like differential geometry, special relativity and theoretical physics in general to express triple products components using tensor notation.

This is because such a representation provides a basis-invariant (or coordinate-independent) way of expressing the properties of the product.

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

Returning to the triple cross product, $${\displaystyle \begin{array}{cc}{\left(\mathbf{a}\times [\mathbf{b}\times \mathbf{c}]\right)}_i& =\left({\delta }_i^{\ell }{\delta }_j^m-{\delta }_i^m{\delta }_j^{\ell }\right)a^j{b}_{\ell }{c}_m\\ & =a^j{b}_i{c}_j-a^j{b}_j{c}_i=b_i(\mathbf{a}\cdot \mathbf{c})-c_i(\mathbf{a}\cdot \mathbf{b}).\end{array}}$$