Cross/Dot Product in n dimensions

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SUMMARY

The equation (c × b) · a = (a × c) · b is explored in the context of R^n, where × denotes the cross product and · denotes the dot product. The cross product is defined only in R^1, R^3, and R^7, making the proof in higher dimensions complex. The discussion suggests utilizing the Triple Product in R^3 and introduces the wedge product as a generalization of the cross product in R^n, resulting in a tensor rather than a vector. The existence of the octonions allows for a cross product definition in R^7, analogous to the quaternion algebra in R^3.

PREREQUISITES
  • Understanding of vector operations in R^n
  • Familiarity with the cross product and dot product
  • Knowledge of the Triple Product in R^3
  • Basic concepts of exterior algebra and wedge products
NEXT STEPS
  • Research the properties and applications of the wedge product in R^n
  • Study the Triple Product identity and its implications in higher dimensions
  • Explore the role of octonions in defining vector operations in R^7
  • Learn about the Hodge dual and its relationship to the cross product in R^3
USEFUL FOR

Mathematicians, physics students, and anyone interested in advanced vector calculus and the properties of vector products in higher-dimensional spaces.

rachbomb
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I would like to show that (cxb).a = (axc).b in Rn where x denotes the cross product and . denotes the dot product.

Since the cross product is only defined in R1, R3, and R7, my inclination is to prove the above equation in cases (case one being a,b,c are vectors in R1, etc). However, this seems a bit tedious, particularly for R7.

I am familiar with the Triple Product in R3, but am unsure if it applies in Rn, and if so how to prove it. If so, this seems like a much quicker and more concise proof.

Please help! Any assistance is greatly appreciated. Thanks!
 
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rachbomb said:
I would like to show that (cxb).a = (axc).b in Rn where x denotes the cross product and . denotes the dot product.

Since the cross product is only defined in R1, R3, and R7,
I've never heard of the cross product being defined in any space other than R3.
rachbomb said:
my inclination is to prove the above equation in cases (case one being a,b,c are vectors in R1, etc). However, this seems a bit tedious, particularly for R7.

I am familiar with the Triple Product in R3, but am unsure if it applies in Rn, and if so how to prove it. If so, this seems like a much quicker and more concise proof.

Please help! Any assistance is greatly appreciated. Thanks!
 
You might want to look into the wedge product (aka exterior algebra) as the generalization of the cross product to R^n. However, the result of the product is a tensor, not a vector. By virtue of the property u \wedge v = - v \wedge u, it satisfies u \wedge v \wedge w = w \wedge u \wedge v. In 3d, you can use the Hodge dual to identify the vector *(u\wedge v) with the cross product and *(u \wedge v \wedge w ) with (u\times v)\cdot w.

To Mark44, it's possible to define a cross product in R^7 because of the existence of the octonions. There's a construction of that product that's analogous to the relationship between the cross product in 3d and the quaternion algebra.
 

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