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

In summary, to find the cross product of two vectors, you can use the rule that ixj= k, jxk= i, kxi= j, and that the cross product is anti-symmetric and linear. In this specific scenario, the cross product of 5k and 3i+4j would be 15j-20i. Another way to calculate this would be to use 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. However, the determinant in this scenario may
  • #1
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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 ?
 
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  • #2
"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
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]
 
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  • #4
HallsofIvy said:
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.
 
  • #5


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.
 

What is the cross product of two vectors?

The cross product of two vectors is a vector that is perpendicular to both of the original vectors and has a magnitude equal to the product of their magnitudes multiplied by the sine of the angle between them.

How is the cross product calculated?

The cross product is calculated by taking the determinant of a 3x3 matrix formed by the two vectors and the unit vectors in the x, y, and z directions. The resulting vector is the cross product.

What is the difference between the dot product and the cross product?

The dot product of two vectors results in a scalar quantity, while the cross product results in a vector quantity. Additionally, the dot product measures the similarity or projection of one vector onto the other, while the cross product measures the perpendicularity between the two vectors.

What are some applications of the cross product?

The cross product is commonly used in physics and engineering, particularly in calculating torque and angular momentum. It is also used in electromagnetic field theory and in computer graphics to determine the orientation of objects in 3D space.

Can the cross product be used in higher dimensions?

The cross product is only defined for three-dimensional vectors. However, there are similar operations for higher dimensions, such as the wedge product in geometric algebra.

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