- #1

- 178

- 4

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 ?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter B4ssHunter
- Start date

- #1

- 178

- 4

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

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 134

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

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

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 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

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]

Last edited by a moderator:

- #4

mathman

Science Advisor

- 7,932

- 484

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 thedeterminant.

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.

Share: