Hello, joepinsley18!
Prove that: . . . . . Not true!
I think you mean: .
We can do it in a most primitive way. .
We have: .
Hence: .
. . . . . . . . . . . . .
We have: .
Hence: .
. . . . . . . . . . . . .
Therefore: .
Prove that (u cross v) dot w = u dot (w cross v)
This is called the triple scalar or box product
Please try it using a parallelpiped but be sure to consider all cases (the angle between the vector normal to the base and the height vector could be obtuse)
I will gladly accept any other ways of proving this statement.
Thank you.
I did intend to prove that "u cross v dot w = u dot w cross v". From your proof, I can see that this statement is false! By switching v and w as you did, the determinants of both matrices become equal. Had you tried proving the above statement, you would have found that the determinants are opposites. Thank you. I see it now.