Cross/Dot Product in n dimensions

[rachbomb]

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!

 

[Mark44]

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.

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!

 

[fzero]

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.