What is the cross product of 5k and 3i+4j?

[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 ?

 

[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.

 

[HallsofIvy]

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
[tex](ai+ bj+ ck)\times (di+ ej+ fk)= \left|\begin{array}{ccc}i & j & k \\ a & b & c \\ d & j & k\end{array}\right|[/tex]
where the right side is the determinant.

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

 

[mathman]

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
[tex](ai+ bj+ ck)\times (di+ ej+ fk)= \left|\begin{array}{ccc}i & j & k \\ a & b & c \\ d & j & k\end{array}\right|[/tex]
where the right side is the determinant.

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

The determinant looks wrong.

 

[CellCoree]

I can confirm that the cross product of 5k and 3i+4j is indeed (15j - 20i). This is because the cross product of two vectors is a vector that is perpendicular to both of the original vectors, and its magnitude is equal to the product of the magnitudes of the two vectors multiplied by the sine of the angle between them. In this case, the angle between 5k and 3i+4j is 90 degrees, making the sine of the angle equal to 1. Therefore, the magnitude of the cross product is equal to the product of the magnitudes of 5k (which is 5) and 3i+4j (which is 5), resulting in a magnitude of 25. The direction of the cross product is determined by the right-hand rule, which in this case, would result in a vector pointing in the direction of -20i+15j.