# Cross product of vectors

1. Sep 27, 2013

### B4ssHunter

i have a vector xK where k is the unit vector perpendicular to other unit vectors i and j
when i multiply a force which has 5k for instance another which has ( 3 i + 4 j )
i multiply 5k by 3i then 5k by 4j right ?
the answer would be ( 15 j - 20 i ) right ?

2. Sep 27, 2013

### arildno

"5k for instance another which has ( 3 i + 4 j )
i multiply 5k by 3i then 5k by 4j right ?"
"the answer would be ( 15 j - 20 i ) right ? "
If your k-vector is the left-hand factor in the cross product, yes.
If your k-vector is your right-hand factor, the signs should be changed.

3. Sep 28, 2013

### HallsofIvy

Staff Emeritus
It's hard to make sense out of this. I presume you are asking specifically about the cross product of the vectors 5k and 3i+ 4j. The basic rules for the cross product is that ixj= k, jxk= i, kxi= j, the cross product is "anti- symmetric" (uxv= -vxu), and linear.

The fact that the cross product is linear means that 5k x(3i+ 4j)= 15 kxi+ 20 kxj. The fact that the cross product is anti-symmetric means that kxj= -jxk= -i, so that 5k x (3i+ 4j)= 15j- 20i.

Most people remember the cross product with the mnemonic
$$(ai+ bj+ ck)\times (di+ ej+ fk)= \left|\begin{array}{ccc}i & j & k \\ a & b & c \\ d & j & k\end{array}\right|$$
where the right side is the determinant.

Here, that would give
$$5k \times 3i+ 4j= \left|\begin{array}{ccc}i & j & k \\ 0 & 0 & 5 \\ 3 & 4 & 0 \end{array}\right|$$
Expanding the determinant on the second row, that is
$$-5\left|\begin{array}{cc}i & j \\ 3 & 4 \end{array}\right|= -5(4i- 3j)= 15j- 20i$$

Last edited: Sep 28, 2013
4. Sep 28, 2013

### mathman

The determinant looks wrong.

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