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Triple product

From Wikipedia, the free encyclopedia
Ternary operation on vectors
This article is about ternary operations on vectors. For other uses, seeTriple product (disambiguation).
"Signed volume" redirects here. For autographed books, seeBibliophilia.

Ingeometry andalgebra, thetriple product is a product of three 3-dimensional vectors, usuallyEuclidean vectors. The name "triple product" is used for two different products, thescalar-valuedscalar triple product and, less often, thevector-valuedvector triple product.

Scalar triple product

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Three vectors defining a parallelepiped

Thescalar triple product (also called themixed product,box product, ortriple scalar product) is defined as thedot product of one of the vectors with thecross product of the other two.

Geometric interpretation

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Geometrically, the scalar triple product

a(b×c){\displaystyle \mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )}

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

Properties

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The triple product is a scalar density

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Strictly speaking, ascalar 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. That is, the triple product of covariant vectors is more properly described as ascalar density.

Ta(Tb×Tc)=det(TaTbTc)=det(T(abc))=det(T)det(abc)=det(T)(a(b×c)){\displaystyle {\begin{aligned}T\mathbf {a} \cdot (T\mathbf {b} \times T\mathbf {c} )&=\det {\begin{pmatrix}T\mathbf {a} &T\mathbf {b} &T\mathbf {c} \end{pmatrix}}\\&=\det \left(T{\begin{pmatrix}\mathbf {a} &\mathbf {b} &\mathbf {c} \end{pmatrix}}\right)\\&=\det(T)\det \!{\begin{pmatrix}\mathbf {a} &\mathbf {b} &\mathbf {c} \end{pmatrix}}\\&=\det(T)\left(\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )\right)\end{aligned}}}

Some authors use "pseudoscalar" to describe an object that looks like a scalar but does not transform like one. Because the triple product transforms as a scalar density not as a scalar, it could be called a "pseudoscalar" by this broader definition. However, the triple product is not a "pseudoscalar density".

When a transformation is an orientation-preserving rotation, its determinant is+1 and the triple product is unchanged. When a transformation is an orientation-reversing rotation then its determinant is−1 and the triple product is negated. An arbitrary transformation could have a determinant that is neither+1 nor−1.

As an exterior product

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The three vectors spanning a parallelepiped have triple product equal to its volume. (However, beware that the direction of the arrows in this diagram is incorrect.)

Inexterior algebra andgeometric algebra the exterior product of two vectors is abivector, while the exterior product of three vectors is atrivector. 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 vectorsa,b andc, the product

abc{\displaystyle \mathbf {a} \wedge \mathbf {b} \wedge \mathbf {c} }

is a trivector with magnitude equal to the scalar triple product, i.e.

|abc|=|a(b×c)|{\displaystyle |\mathbf {a} \wedge \mathbf {b} \wedge \mathbf {c} |=|\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )|},

and is theHodge dual of the scalar triple product. As the exterior product is associative brackets are not needed as it does not matter which ofab orbc is calculated first, though the order of the vectors in the product does matter. Geometrically the trivectorabc corresponds to the parallelepiped spanned bya,b, andc, with bivectorsab,bc andac matching theparallelogram faces of the parallelepiped.

As a trilinear function

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The triple product is identical to thevolume form of the Euclidean 3-space applied to the vectors viainterior product. It also can be expressed as acontraction of vectors with a rank-3 tensor equivalent to the form (or apseudotensor equivalent to the volume pseudoform); seebelow.

Vector triple product

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

a×(b×c)=(ac)b(ab)c{\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 astriple product expansion, orLagrange's formula,[2][3] although the latter name is also used forseveral other formulas. Its right hand side can be remembered by using themnemonic "ACB − ABC", provided one keeps in mind which vectors are dotted together. A proof is providedbelow. Some textbooks write the identity asa×(b×c)=b(ac)c(ab){\displaystyle \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 familiarmnemonic "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:

(a×b)×c=c×(a×b)=(cb)a+(ca)b{\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:

a×(b×c)+b×(c×a)+c×(a×b)=0{\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 theJacobi identity for the cross product. Another useful formula follows:

(a×b)×c=a×(b×c)b×(a×c){\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 inphysics. A related identity regardinggradients and useful invector calculus is Lagrange's formula of vector cross-product identity:[4]

×(×A)=(A)()A{\displaystyle {\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 generalLaplace–de Rham operatorΔ=dδ+δd{\displaystyle \Delta =d\delta +\delta d}.

Proof

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Thex{\displaystyle x} component ofu×(v×w){\displaystyle \mathbf {u} \times (\mathbf {v} \times \mathbf {w} )} is given by:

(u×(v×w))x=uy(vxwyvywx)uz(vzwxvxwz)=vx(uywy+uzwz)wx(uyvy+uzvz)=vx(uywy+uzwz)wx(uyvy+uzvz)+(uxvxwxuxvxwx)=vx(uxwx+uywy+uzwz)wx(uxvx+uyvy+uzvz)=(uw)vx(uv)wx{\displaystyle {\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, they{\displaystyle y} andz{\displaystyle z} components ofu×(v×w){\displaystyle \mathbf {u} \times (\mathbf {v} \times \mathbf {w} )} are given by:

(u×(v×w))y=(uw)vy(uv)wy(u×(v×w))z=(uw)vz(uv)wz{\displaystyle {\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:

u×(v×w)=(uw) v(uv) w{\displaystyle \mathbf {u} \times (\mathbf {v} \times \mathbf {w} )=(\mathbf {u} \cdot \mathbf {w} )\ \mathbf {v} -(\mathbf {u} \cdot \mathbf {v} )\ \mathbf {w} }[5]

Using geometric algebra

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If geometric algebra is used the cross productb ×c of vectors is expressed as their exterior productbc, abivector. The second cross product cannot be expressed as an exterior product, otherwise the scalar triple product would result. Instead aleft contraction[6] can be used, so the formula becomes[7]

a(bc)=b(ac)(ab)c=(ac)b(ab)c{\displaystyle {\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.[6] The result is the same vector as calculated usinga × (b ×c).

Triple bivector product

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

a×(b×c)=(ac)b+(ab)c{\displaystyle {\overset {\Rightarrow }{a}}\times \left({\overset {\Rightarrow }{b}}\times {\overset {\Rightarrow }{c}}\right)=-\left({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{c}}\right){\overset {\Rightarrow }{b}}+\left({\overset {\Rightarrow }{a}}\cdot {\overset {\Rightarrow }{b}}\right){\overset {\Rightarrow }{c}}}

Proof

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

(a(bc))=12(a(bc)(bc)a)=a×(bc)(a×(bc))=12(a12(bccb)12(bccb)a)=a×(bc)(a×(bc))=12(a12(bccb)12(bccb)a)=a×(b×c){\displaystyle {\begin{alignedat}{3}&(-\mathbf {a} \;{\big \lrcorner }\;(\mathbf {b} \wedge \mathbf {c} ))\star &&=-{\tfrac {1}{2}}\left({\overset {\Rightarrow }{a}}(\mathbf {b} \wedge \mathbf {c} )-(\mathbf {b} \wedge \mathbf {c} ){\overset {\Rightarrow }{a}}\right)&&=-{\overset {\Rightarrow }{a}}\times (\mathbf {b} \wedge \mathbf {c} )\\&(-{\overset {\Rightarrow }{a}}\times (\mathbf {b} \wedge \mathbf {c} ))\star &&=-{\tfrac {1}{2}}\left({\overset {\Rightarrow }{a}}{\tfrac {1}{2}}({\overset {\Rightarrow }{b}}\mathbf {c} -\mathbf {c} {\overset {\Rightarrow }{b}})-{\tfrac {1}{2}}({\overset {\Rightarrow }{b}}\mathbf {c} -\mathbf {c} {\overset {\Rightarrow }{b}}){\overset {\Rightarrow }{a}}\right)&&=-{\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\cdot \mathbf {c} )\\&(-{\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\cdot \mathbf {c} ))\star &&=-{\tfrac {1}{2}}\left({\overset {\Rightarrow }{a}}{\tfrac {1}{2}}({\overset {\Rightarrow }{b}}{\overset {\Rightarrow }{c}}-{\overset {\Rightarrow }{c}}{\overset {\Rightarrow }{b}})-{\tfrac {1}{2}}({\overset {\Rightarrow }{b}}{\overset {\Rightarrow }{c}}-{\overset {\Rightarrow }{c}}{\overset {\Rightarrow }{b}}){\overset {\Rightarrow }{a}}\right)&&={\overset {\Rightarrow }{a}}\times ({\overset {\Rightarrow }{b}}\times {\overset {\Rightarrow }{c}})\end{alignedat}}}

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

((((ac)b(ab)c)))=12(ac+ca)b12(ab+ba)c=(ac)b+(ab)c{\displaystyle {\begin{aligned}&((((\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} )\star )\star )\star \\&={\tfrac {1}{2}}({\overset {\Rightarrow }{a}}{\overset {\Rightarrow }{c}}+{\overset {\Rightarrow }{c}}{\overset {\Rightarrow }{a}}){\overset {\Rightarrow }{b}}-{\tfrac {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:

a×(b×c)=(ac)b+(ab)c{\displaystyle {\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}}}

Triple products using tensor notation

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It can be useful in fields likedifferential geometry,special relativity andtheoretical physics in general to express triple products components usingtensor notation.

This is because such a representation provides abasis-invariant (orcoordinate-independent) way of expressing the properties of the product.

The triple scalar product is expressed using theLevi-Civita symbol:[8]a[b×c]=εijkaibjck{\displaystyle \mathbf {a} \cdot [\mathbf {b} \times \mathbf {c} ]=\varepsilon _{ijk}a^{i}b^{j}c^{k}}while the triple vector product:(a×[b×c])i=εijkajεkmbcm=εijkεkmajbcm,{\displaystyle (\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 thei{\displaystyle i}-th component of the resulting vector. This can be simplified by performing acontraction on theLevi-Civita symbols,εijkεkm=δijm=δiδjmδimδj,{\displaystyle \varepsilon _{ijk}\varepsilon ^{k\ell m}=\delta _{ij}^{\ell m}=\delta _{i}^{\ell }\delta _{j}^{m}-\delta _{i}^{m}\delta _{j}^{\ell }\,,}whereδji{\displaystyle \delta _{j}^{i}} is theKronecker delta function (δji=0{\displaystyle \delta _{j}^{i}=0} whenij{\displaystyle i\neq j} andδji=1{\displaystyle \delta _{j}^{i}=1} wheni=j{\displaystyle i=j}) andδijm{\displaystyle \delta _{ij}^{\ell m}} is thegeneralized Kronecker delta function. We can reason out this identity by recognizing that the indexk{\displaystyle k} will be summed out leaving onlyi{\displaystyle i} andj{\displaystyle j}. In the first term, we fixi={\displaystyle i=\ell } and thusj=m{\displaystyle j=m}. Likewise, in the second term, we fixi=m{\displaystyle i=m} and thus=j{\displaystyle \ell =j}.

Returning to the triple cross product,(a×[b×c])i=(δiδjmδimδj)ajbcm=ajbicjajbjci=bi(ac)ci(ab).{\displaystyle {\begin{aligned}\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}\\[1ex]&=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{aligned}}}

See also

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Notes

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  1. ^Wong, Chun Wa (2013).Introduction to Mathematical Physics: Methods & Concepts. Oxford University Press. p. 215.ISBN 9780199641390.
  2. ^Joseph Louis Lagrange did not develop the cross product as an algebraic product on vectors, but did use an equivalent form of it in components: seeLagrange, J-L (1773). "Solutions analytiques de quelques problèmes sur les pyramides triangulaires".Oeuvres. Vol. 3. He may have written a formula similar to the triple product expansion in component form. See alsoLagrange's identity andKiyosi Itô (1987).Encyclopedic Dictionary of Mathematics. MIT Press. p. 1679.ISBN 0-262-59020-4.
  3. ^Kiyosi Itô (1993)."§C: Vector product".Encyclopedic dictionary of mathematics (2nd ed.). MIT Press. p. 1679.ISBN 0-262-59020-4.
  4. ^Pengzhi Lin (2008).Numerical Modelling of Water Waves: An Introduction to Engineers and Scientists. Routledge. p. 13.ISBN 978-0-415-41578-1.
  5. ^J. Heading (1970).Mathematical Methods in Science and Engineering. American Elsevier Publishing Company, Inc. pp. 262–263.
  6. ^abPertti Lounesto (2001).Clifford algebras and spinors (2nd ed.). Cambridge University Press. p. 46.ISBN 0-521-00551-5.
  7. ^Janne Pesonen."Geometric Algebra of One and Many Multivector Variables"(PDF). p. 37.
  8. ^"Permutation Tensor". Wolfram. Retrieved21 May 2014.

References

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  • Lass, Harry (1950).Vector and Tensor Analysis. McGraw-Hill Book Company, Inc. pp. 23–25.

External links

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Linear equations
Three dimensional Euclidean space
Matrices
Matrix decompositions
Relations and computations
Vector spaces
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Multilinear algebra
Affine and projective
Numerical linear algebra
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