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tib

It is correct that the determinant method would only work if the vectors lie in 3D space?

keenan

The determinant is well-defined in any dimension, but formulas involving the cross product will work only in 3D.

Azure

I don't think this is true; there should be more than one vector that satisfies this. Considering that u x v is the actual cross product, since the determinant is a multilinear function, and det(u,v,v) = 0, then det(u,v,uxv) = det(u,v,uxv)+0 = det(u,v,uxv) + det(u,v,v) = det(u,v,uxv+v), constructing another vector that has the same requirements.

I think that in order to truly limit it to one vector, we also have to put in the conditions that u.dot(u.cross(v)) = v.dot(u.cross(v)) = 0... which still doesn't work if v and u are parallel.

triangle

Based on this unique formula, we could also express this cross product u x v by using a matrix to represent the coefficients in u and multiplying by the vector v.

small_potato__

Fun fact: earlier versions of Matlab allowed you to take a cross product in multiple dimensions, but in recent updates, they have taken this away and made it only for vectors of length 3.