2019年3月22日 · E. A. Abbott describes a 2D cross product nicely in his mathematical fantasy book "Flatland": Flatland describes life and customs of people in a 2-D world: in this universe vectors can be summed together and projected, areas are calculated, rotations are clock-wise or counter clock-wise, reflection is possible...

2017年2月4日 · The main thing is that my professor asked for us to solve this without using the "determinant method." I have just started linear algebra, so I am still trying to understand determinants and the like. I am just wondering how it is possible to prove the cross product of two vectors with another method? Any help would be great!

So the answer to your question is that the cross product of two parallel vectors is $\mathbf 0$ because the rejection of a vector from a parallel vector is $\mathbf 0$ and hence has length $0$. Share Cite

2022年7月16日 · Therefore, the cross product of the two normal vectors will be parallel to each of the two planes. Which means it will also be parallel to the common line shared by the two planes - their line of intersection. Proving it mathematically in your specific case is quite trivial, and you should be able to do it by following definitions. Just find ...

Look. I'm not a mathematician, but I have a perspective which can explain why the cross product of two vectors is another vector perpendicular to them. It is not a proof but it will help make that idea fimilar. One can understand cross product in this way:imagine a line segment that makes colorful marks wherever it moves on a paper.

Thus, the cross product of two unit vectors $\vec{u}$ and $\vec{v}$ is itself a unit vector if and only if $\vec{u}$ and $\vec{v}$ are orthogonal, i.e. meet at right angles (this makes $\sin(\theta)=\sin(\frac{\pi}{2})=1$). As to the general interpretation of the magnitude of the cross product, see Wikipedia:

2016年2月15日 · The difference between an ordered pair of vectors and a tensor product of vectors is that if you multiply one of the vectors by a nonzero scalar and the other by the reciprocal of that scalar, then you get a different ordered pair but the same tensor product. $\qquad$ $\endgroup$ –

Two more uses that I haven't seen mentioned yet: if you want to find the area of the parallelogram formed by two vectors (each vector gives a pair of parallel sides), then you would use the magnitude of the cross product of the two vectors. One use of this is to aid in defining a surface integral. Let x(u,v) be a parameterization of a surface.

2023年12月28日 · You might have two vectors whose cross product is $(5, 3, 2)$ under regular coordinates, but if you changed your coordinate system to switch the first and second dimensions, without (anti)symmetry the cross product could have an entirely different value, like $(-1, 4, 1)$. A mathematical operation that depends on something totally unphysical ...

2015年3月13日 · $\begingroup$ The meaning of triple product (x × y)⋅ z of Euclidean 3-vectors is the volume form (SL(3, ℝ) invariant), that gets an expression through dot product (O(3) invariant) and cross product (SO(3) invariant, a subgroup of SL(3, ℝ)). We can complexify all the stuff (resulting in SO(3, ℂ)-invariant vector calculus), although we ...