In mathematics, the quadruple product is a product of four vectors in three-dimensional Euclidean space. The name "quadruple product" is used for two different products,[1] the scalar-valued scalar quadruple product and the vector-valued vector quadruple product or vector product of four vectors.
Scalar quadruple product
The scalar quadruple product is defined as the dot product of two cross products:
![{\displaystyle (\mathbf {a\times b} )\cdot (\mathbf {c} \times \mathbf {d} )\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee68c7cbb48abac3882d1406b384a472bad3bb4c)
where a, b, c, d are vectors in three-dimensional Euclidean space.[2] It can be evaluated using the identity:[2]
![{\displaystyle (\mathbf {a\times b} )\cdot (\mathbf {c} \times \mathbf {d} )=(\mathbf {a\cdot c} )(\mathbf {b\cdot d} )-(\mathbf {a\cdot d} )(\mathbf {b\cdot c} )\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49d8dac362692f03e11cde6f23661c15aac1edd4)
or using the determinant:
![{\displaystyle (\mathbf {a\times b} )\cdot (\mathbf {c} \times \mathbf {d} )={\begin{vmatrix}\mathbf {a\cdot c} &\mathbf {a\cdot d} \\\mathbf {b\cdot c} &\mathbf {b\cdot d} \end{vmatrix}}\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bce6a5126673f1d21c35b7f901bc1ba6b1797d1)
Proof
We first prove that
![{\displaystyle {\begin{aligned}\mathbf {c} \times (\mathbf {b} \times \mathbf {a} )\cdot \mathbf {d} =(\mathbf {a} \times \mathbf {b} )\cdot (\mathbf {c} \times \mathbf {d} ).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f17e30884d8cb7c0a0e067a92fd96d54d497527)
This can be shown by straightforward matrix algebra using the correspondence between elements of
and
, given by
, where
![{\displaystyle {\begin{aligned}\mathbf {\hat {a}} ={\begin{bmatrix}0&-a_{3}&a_{2}\\a_{3}&0&-a_{1}\\-a_{2}&a_{1}&0\end{bmatrix}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28b982a3fbd0cd9ceec69499a5271a475fd88030)
It then follows from the properties of skew-symmetric matrices that
![{\displaystyle {\begin{aligned}\mathbf {c} \times (\mathbf {b} \times \mathbf {a} )\cdot \mathbf {d} =(\mathbf {\hat {c}} \mathbf {\hat {b}} \mathbf {a} )^{\mathrm {T} }\mathbf {d} =\mathbf {a} ^{\mathrm {T} }\mathbf {\hat {b}} \mathbf {\hat {c}} \mathbf {d} =(-\mathbf {\hat {b}} \mathbf {a} )^{\mathrm {T} }\mathbf {\hat {c}} \mathbf {d} =(\mathbf {\hat {a}} \mathbf {b} )^{\mathrm {T} }\mathbf {\hat {c}} \mathbf {d} =(\mathbf {a} \times \mathbf {b} )\cdot (\mathbf {c} \times \mathbf {d} ).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c5fbf3e05b521b848dbe8eb7bbc4ef933b01196)
We also know from vector triple products that
![{\displaystyle {\begin{aligned}\mathbf {c} \times (\mathbf {b} \times \mathbf {a} )=(\mathbf {c} \cdot \mathbf {a} )\mathbf {b} -(\mathbf {c} \cdot \mathbf {b} )\mathbf {a} .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94f592bd4c324724fbadc6005bea1d1b23fed705)
Using this identity along with the one we have just derived, we obtain the desired identity:[citation needed]
![{\displaystyle {\begin{aligned}(\mathbf {a} \times \mathbf {b} )\cdot (\mathbf {c} \times \mathbf {d} )=\mathbf {c} \times (\mathbf {b} \times \mathbf {a} )\cdot \mathbf {d} =\left[(\mathbf {c} \cdot \mathbf {a} )\mathbf {b} -(\mathbf {c} \cdot \mathbf {b} )\mathbf {a} \right]\cdot \mathbf {d} =(\mathbf {a} \cdot \mathbf {c} )(\mathbf {b} \cdot \mathbf {d} )-(\mathbf {a} \cdot \mathbf {d} )(\mathbf {b} \cdot \mathbf {c} ).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a789aecc44d71c52b06ba6dbd684872d541766b)
Vector quadruple product
The vector quadruple product is defined as the cross product of two cross products:
![{\displaystyle (\mathbf {a\times b} )\mathbf {\times } (\mathbf {c} \times \mathbf {d} )\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/931eb31f433fe684794b0f1725ac2273feba3237)
where a, b, c, d are vectors in three-dimensional Euclidean space.[3] It can be evaluated using the identity:[4]
![{\displaystyle (\mathbf {a\times b} )\mathbf {\times } (\mathbf {c} \times \mathbf {d} )=[\mathbf {a,\ b,\ d} ]\mathbf {c} -[\mathbf {a,\ b,\ c} ]\mathbf {d} \ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83e41e3db9c5e3a2fb2030f9822ab0af1f24ad91)
using the notation for the triple product:
![{\displaystyle [\mathbf {a,\ b,\ c} ]=\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aef3f5f3c5d0bb976d5ae3b8e1ccd43861fdbc11)
Equivalent forms can be obtained using the identity:[5]
![{\displaystyle [\mathbf {b,\ c,\ d} ]\mathbf {a} -[\mathbf {c,\ d,\ a} ]\mathbf {b} +[\mathbf {d,\ a,\ b} ]\mathbf {c} -[\mathbf {a,\ b,\ c} ]\mathbf {d} =0\ .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6aab120a97d92e8ffd4c390fc8be37ebdf564021)
This identity can also be written using tensor notation and the Einstein summation convention as follows:
![{\displaystyle (\mathbf {a\times b} )\mathbf {\times } (\mathbf {c} \times \mathbf {d} )=\varepsilon _{ijk}a^{i}c^{j}d^{k}b^{l}-\varepsilon _{ijk}b^{i}c^{j}d^{k}a^{l}=\varepsilon _{ijk}a^{i}b^{j}d^{k}c^{l}-\varepsilon _{ijk}a^{i}b^{j}c^{k}d^{l}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d5f46c7801684fd548b8cc7e821140e33b1c485)
Application
The quadruple products are useful for deriving various formulas in spherical and plane geometry.[3] For example, if four points are chosen on the unit sphere, A, B, C, D, and unit vectors drawn from the center of the sphere to the four points, a, b, c, d respectively, the identity:
![{\displaystyle (\mathbf {a\times b} )\mathbf {\cdot } (\mathbf {c\times d} )=(\mathbf {a\cdot c} )(\mathbf {b\cdot d} )-(\mathbf {a\cdot d} )(\mathbf {b\cdot c} )\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80a90640b0c3ba354b957793f93be2695fe66b15)
in conjunction with the relation for the magnitude of the cross product:
![{\displaystyle \|\mathbf {a\times b} \|=ab\sin \theta _{ab}\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e484cc61363a8c51ecc133887e61cea4afb28887)
and the dot product:
![{\displaystyle \mathbf {a\cdot b} =ab\cos \theta _{ab}\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4debac0acc72d3eee5263089daf49336b3675f92)
where a = b = 1 for the unit sphere, results in the identity among the angles attributed to Gauss:
![{\displaystyle \sin \theta _{ab}\sin \theta _{cd}\cos x=\cos \theta _{ac}\cos \theta _{bd}-\cos \theta _{ad}\cos \theta _{bc}\ ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68d9e5af6ab0b7b37e3d56cee36faefded04ecef)
where x is the angle between a × b and c × d, or equivalently, between the planes defined by these vectors.
Josiah Willard Gibbs's pioneering work on vector calculus provides several other examples.[3]
Notes
References
See also