# Product of Vectors (cross)

1. Sep 13, 2009

### Neophyte

1. The problem statement, all variables and given/known data

^ ^ ^
i = i; j = j; k = k;

i = x j = y k = z
So this is the result after cross multiplication.

-2ij - 8ji = 6k

-ik + 4ki = 5j

-4jk +4kj = -8i

Unfortunately I am a little confused with the 5j
It seems to me that it would be

-ik + 4ki = -j -4j

because I comes first and so it would be -5j

I know this is not the case but it ruins the logic of the whole ordeal for me ; /. How are you supposed to know which sign you need to flip? I thought it was if it was ij you would leave it alone but if it was ji it would be -ji because order changed and i is multiplied first.

2. Sep 13, 2009

### rl.bhat

Cross multiplication of which vectors?

3. Sep 13, 2009

### Neophyte

It is a . (b x c)
http://img7.imageshack.us/img7/7537/problemd.th.png [Broken]

Basically if

a x b = (aybz - byaz)i + (azbx - bzax)j + (axby - bxay)k

I assumed that the second part was negative because a x b = -b x a
and so (aybz - byaz)i + (axby - bxay)k seemed alright but then the (azbx - bzax)j messed everything up for me because I thought it would be (axbz - bzax)j.
Then after that idea failed miserably I had no clue. I have memorized the end result at this point but I havent a clue how to get there ; (

Last edited by a moderator: May 4, 2017
4. Sep 13, 2009

### rl.bhat

-ik + 4ki = -j -4j
ixk = -j and kxi = j
so -(-j) + 4J = 5J

5. Sep 13, 2009

### Neophyte

But what makes the ixk change signs as oppose to the kxi ?
I am missing how the kxi takes precedence over the ixk. Wouldn't it be the other way around?
You multiple the ixk first and i comes before k in equation :(.

Is there a rule to determine which flips?

6. Sep 13, 2009

Yes.
AxB = - BxA