# Homework Help: Help with cross product proofs

1. Sep 1, 2008

### imsoconfused

I have a 23 problem assignment due at the end of the week, and although I'm going to have a chance to talk to my teacher about the questions I have, I'd like to go ahead and get going on the problems. I've successfully completed 21 of them, but the last two are stumping me. I'm submitting them here for your help. I really just don't understand!!!

1. split (A+B)x(A+B) into four terms to deduce that (AXB)=-(BXA)
my first thought is that I need to expand them (obviously), but I'm not quite sure exactly what to expand them TO. I have tried making it (A+B)A X (A+B)B, but that yields (A^2+AB) X (AB+B^2) and I'm not sure how much good that does me. If you could just point me in the direction of what to expand, I think I can get it from there.

2. verify B.(AXB)=0
um... I have absolutely no clue as to what to do. do I use arbitrary variables and solve (e.g. use A=<x,y,z>)?

I know this is a lot to ask, but I just want to get ahead on this homework and understand generally what I'm doing before I talk to my teacher.

2. Sep 2, 2008

### Defennder

The 2nd question is related to the first. For the 2nd, just think of any vector at all and compute vXv. When you have that, apply the property (not that particular result!) to 1. To prove 1, just expand the expression out using the distributive property of the cross product.

3. Sep 2, 2008

### HallsofIvy

What do you mean by "AB"? Remember that there are two different kinds of multiplication for vectors, dot product and cross product. Don't forget to write the "X"!

More importantly, (A+B)X(A+ B) is NOT "(A+B)XA X (A+B)XB", it is "(A+B)XA + (A+B)XB". Now expand each of those. Are you allowed to use VXV= 0 for any vector V?

4. Sep 2, 2008

### imsoconfused

thank you! hallsofivy, you're so right!! if I hadn't been so careless with my signs, I would have seen that this is a really easy problem. I finished the first one, and now I have an idea as to how to do the second. I have to leave the computer right now, but I will come back and post what I got.
thanks again!

5. Sep 2, 2008

### imsoconfused

ok, I feel stupid for asking that second question. I ended up with the correct answer (checked against several of my classmates) for both, so I'm confident I understand what I need to do from here. sorry for being retarded! =P