i, j, k unit vectors

[PG1995]

Hi

I'm little confused about the operations on unit vectors i, j, k which are in the directions of positive x, y, z axes. My confusion is about the #2 and #3. For instance, #2 says i x j = 1 but #3 says i x j = k. What's the difference? Is #2 giving information about the magnitude and #3 about the direction? Are there some other related operations which can be included here? Please help me with it. Thanks a lot.

1: i . i = 1 ; i . j = 0 ; i . k = 0

2: i x i = 0 ; i x j = 1 ; i x k = 1

3: i x j = k ; j x k = i ; k x i = j

Regards
PG

 

[misterT]

The result of a cross product is always a vector or a zero vector. So there must be an error (typo) in the text.
- Lines 1 and 3 are correct.
- i x i = 0 is correct.
- These are incorrect: i x j = 1 ; i x k = 1

 

[PG1995]

Thank you, misterT.

Regards
PG

 

[MrAl]

Hi,

Yes something is definitely wrong with line 2. It's not even the magnitude in that last one it's totally wrong.

 

[steveB]

Hi

I'm little confused about the operations on unit vectors i, j, k which are in the directions of positive x, y, z axes. My confusion is about the #2 and #3. For instance, #2 says i x j = 1 but #3 says i x j = k. What's the difference? Is #2 giving information about the magnitude and #3 about the direction? Are there some other related operations which can be included here? Please help me with it. Thanks a lot.

1: i . i = 1 ; i . j = 0 ; i . k = 0

2: i x i = 0 ; i x j = 1 ; i x k = 1

3: i x j = k ; j x k = i ; k x i = j

Regards
PG

I'm not sure what source you found this in, but I suspect the following is the intention of the author. Simply add the magnitude operator to line 2.

1: i . i = 1 ; i . j = 0 ; i . k = 0

2: | i x i| = 0 ; | i x j| = 1 ; | i x k| = 1

3: i x j = k ; j x k = i ; k x i = j

 

[MrAl]

Hi Steve,

Yes that's what i was thinking but then the -1 in the last result threw me a curve.

 

[steveB]

Hi Steve,

Yes that's what i was thinking but then the -1 in the last result threw me a curve.

Yes, and the magnitude of -1 is |-1|=1. So, all is good.

 

[PG1995]

Thanks a lot, MrAl, Steve.

With best wishes
PG