View Full Version : vector triple product proof

04-27-2007, 06:21 PM
Im trying to prove this identity.

Let u v and w be vectors in (Rn) and <u,u> denote the dot product.

u x ( v x w ) = <u,w> v - <u,v> w

Here are my ideas on this.

I tried using the normal algebraic properties of the cross product... dead end.

My next idea is to just prove it for (R3)

Let u = (u1, u2, u3) v= w = and so on, and just plug it in on the left side and try to find the right.

If that does work, could I assume that it will work for (Rn), becuase pluging in u = (u1, u2, u3...un) for each vector seems crazy.

Any ideas or other ways to prove this one? Not looking for the proof just a place to start.


04-27-2007, 07:10 PM
This is an awful, awful problem to do.
There is really only one way to do it.
Here it is: give each vector a set of coordinates.
Then do the cross product operations: that is truly awful.
Then you have a real job of factoring ahead of you.
You can find this done in a good calculus textbook such as Stewart or Thomas.

04-27-2007, 07:48 PM
thats how I did it... for u v w being ordered triplets. (R3)

But does that truely prove that it works for (Rn)?

Or do i need to do it somehow like this...

u = (u1, u2, u3, ... un) which seems rediculous becuase the nth U wouldn't neccesarly match with the determinates sign change easly.

I think I just going to stick with u = (u1, u2, u3)

one more question though.

If I work the identities on both side and meet in the middle, would that be accepted as proof by most?

04-27-2007, 08:05 PM
Does this have a generlization to n-space?
Basicaly ux(vxw) is a plane, determined v & w.
There may be a similar concept of a hyper-plane? I don't know.

04-28-2007, 05:53 AM
I think I missed something before... they book probably is refering to 3 space. I was just asumming for some reason that the vector triple product was used for vectors larger than ordered triplets. But proving it for three space seems the only logical way... especially for a Intro to Applied Linear Algebra class.