Cross Product Derivative Proof

[I Love Math]

Hello again. I'm in need of a push in the right direction here. I have to prove the theorem that (v x w)' = (v x w') + (v' x w). I worked out both sides of the equation, but I've been told that you have to stick with one side like a Trig identity.

I said that v = ai+bj+ck and w = xi+yj+zk then I took the cross product to get (bz-cy)i - (az-cx)j + (ay-bx)k. (Definition of Cross Product)

Then I took the derivative to get:

[(bz'+b'z)-(cy'+c'y)]i - [(az'+a'z)-(cx'+c'x)]j + [(ay'+a'y)-(bx'+b'x)]k
(Power Rule for Derivatives)

I distributed the minus signs:

(bz'+b'z-cy'-c'y)i - (az'+a'z-cx'-c'x)j + (ay'+a'y-bx'-b'x)k

Then I started working on the other side:

(v x w') + (v' x w)

I took the derivative of w: w' = x'i + y'j + z'k
(Derivative of a Vector)

I crossed that with v:

(bz'-cy')i - (az'-cx')j + (ay'-bx')k
(Definition of Cross Product)

Then, I took the derivative of v: v' = a'i + b'j + c'k
(Derivative of a Vector)

I crossed that with w:

(b'z-c'y)i - (a'z-c'x)j + (a'y-b'x)k
(Definition of a Cross Product)

Then, I added (v x w') + (v' x w):

(bz'+b'z-cy'-c'y)i - (az'+a'z-cx'-c'x)j + (ay'+a'y-bx'-b'x)k
(Vector Addition)

This is the same as the result from (v x w)', but I've been told that's not how a proof works. Do I get to this:

(v x w)' = (bz'+b'z-cy'-c'y)i - (az'+a'z-cx'-c'x)j + (ay'+a'y-bx'-b'x)k

and start working backwards? I haven't had to do proofs since high school Geometry class and I'm not totally sure how to work them. Any insight would be greatly appreciated!

Oh, and please don't just post the answer. I really only want a hint.

Thanks!!!

 

[pka]

Why do you think that you have not done a proof?
All of your reasoning is correct.
Some instructors do insist that proof be ‘one-sided’.
In that case, take your ‘second side’ and turn it on its head.

 

[I Love Math]

Thanks for the reply! I will definitely go with this, then!

You guys are the best! )