- #1
Odious Suspect
- 43
- 0
If the cross product in ℝ3 is defined as the area of the parallelogram determined by the constituent vectors joined at the tail, how does one go about proving this product to distribute over vector addition?
I've attached a drawing showing cyan x yellow, cyan x magenta, and cyan x (magenta + yellow). If the parallelogram cyan x yellow is parallel transported along the magenta vector, an open-ended prism is formed, with the these cross products as faces.
If the open ends are capped by congruent triangles and the vector sum of all faces, treated as surface elements with outward-facing normals is taken, the result is the null vector. Since the end-caps cancel, this tells me the sum of any two of the remaining faces is equal to the other.
I believe this result, somehow demonstrates the proposition, but I don't know if it would persuade an independent observer.
My biggest concern is that the general proof that the sum of outward facing surface elements is the null vector relies on the distributive property of the cross product.
I've attached a drawing showing cyan x yellow, cyan x magenta, and cyan x (magenta + yellow). If the parallelogram cyan x yellow is parallel transported along the magenta vector, an open-ended prism is formed, with the these cross products as faces.
If the open ends are capped by congruent triangles and the vector sum of all faces, treated as surface elements with outward-facing normals is taken, the result is the null vector. Since the end-caps cancel, this tells me the sum of any two of the remaining faces is equal to the other.
I believe this result, somehow demonstrates the proposition, but I don't know if it would persuade an independent observer.
My biggest concern is that the general proof that the sum of outward facing surface elements is the null vector relies on the distributive property of the cross product.