Cross Product of Two Vectors

In summary, the conversation discusses the result of applying the cross product to two vectors, A and B, and finding the magnitude of the resulting vector, C. It is shown that the magnitudes of C and 2(AxB) are equal, and the unit vectors for C and (AxB) are also equal. This is because the cross product changes sign when the order of the vectors is changed, but the magnitude remains the same. Additionally, the magnitude of a unit vector is always 1.
  • #1
Mosaness
92
0
1. See attached image please!
2. For part (a), I applied the cross product and got (-6i - 2k) for ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex]. I got (6i + 2k) for ([itex]\vec{B}[/itex] x [itex]\vec{A}[/itex]).

For part (b), [itex]\vec{C}[/itex] was simply (-6i - 2k) - (6i + 2k) = (-12i -4k).

For part (c), the magnitude of [itex]\vec{C}[/itex] was simply 12.65 and for the magnitude of two times ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]) is 12.65. So they are equal. But WHY? I can prove it mathematically, but I'm having some trouble with this.

I do think that it is because ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex] is equal in magnitude but opposite in direction to ([itex]\vec{B}[/itex] x [itex]\vec{A}[/itex]), therefore, the magnitude for 2 times ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]) ought to equal the magnitude of ([itex]\vec{C}[/itex])
 

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  • #2
Mosaness said:
1. See attached image please!
2. For part (a), I applied the cross product and got (-6i - 2k) for ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex]. I got (6i + 2k) for ([itex]\vec{B}[/itex] x [itex]\vec{A}[/itex]).

For part (b), [itex]\vec{C}[/itex] was simply (-6i - 2k) - (6i + 2k) = (-12i -4k).

For part (c), the magnitude of [itex]\vec{C}[/itex] was simply 12.65 and for the magnitude of two times ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]) is 12.65. So they are equal. But WHY? I can prove it mathematically, but I'm having some trouble with this.

I do think that it is because ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex] is equal in magnitude but opposite in direction to ([itex]\vec{B}[/itex] x [itex]\vec{A}[/itex]), therefore, the magnitude for 2 times ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]) ought to equal the magnitude of ([itex]\vec{C}[/itex]) (

You are right the cross product changes sign when you change the order of the vectors, but the magnitude stays the same. Think how the cross product was defined: AxB is a vector perpendicular to both A and B and it points in the direction from where the rotation of the first vector into the second looks anti-clockwise. So AxB=P and BxA=-P. If you subtract -P it is the same as adding P.ehild
 
  • #3
I was doing part (d.) and the unit vector for [itex]\vec{C}[/itex] was (-0.949i - 0.316k) and the unit vector for ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]) was also (-0.949i - 0.316k). Therefore, the unit vector for [itex]\vec{C}[/itex] is not twice as long as the unit vector for ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]). Instead, it is equal. Why is it equal? I'm not quite sure. But if I had to guess, I would say that for vector C, the vector was twice that of ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]), as was the magnitude. And for ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]), the vector and magnitude for half of that for vector C, therefore, when the unit vector was found, they were equal to one another. Had the magnitude of ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]) been half that of vector C, THEN the unit vector for ([itex]\vec{A}[/itex] x [itex]\vec{B}[/itex]) would have been half that of vector C.
 
  • #4
The magnitude of a unit vector is 1. It is the definition of the unit vector: a vector pointing in a specific direction, and having unit length (magnitude).

ehild
 
  • #5
Uh, I think you're overthinking this. What is the magnitude of any unit vector?
 
  • #6
Well the magnitude will always be one. I WAS over thinking it! Oops
 

1. What is the cross product of two vectors?

The cross product of two vectors is a mathematical operation that produces a new vector that is perpendicular to both of the original vectors. It is also known as the vector product, and is denoted by the symbol "x".

2. How is the cross product calculated?

The cross product of two vectors, A and B, can be calculated using the following formula:
A x B = (AyBz - AzBy)i + (AzBx - AxBz)j + (AxBy - AyBx)k
where i, j, and k are unit vectors in the x, y, and z directions respectively.

3. What is the magnitude of the cross product?

The magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by the two vectors. It can be calculated using the formula:
|A x B| = |A||B|sin(θ)
where |A| and |B| are the magnitudes of the two vectors and θ is the angle between them.

4. What is the direction of the cross product?

The direction of the cross product is perpendicular to the plane formed by the two original vectors. The right-hand rule is often used to determine the direction of the resulting vector, where the thumb points in the direction of the cross product when the fingers of the right hand are curled in the direction from the first vector to the second vector.

5. What are the applications of the cross product?

The cross product has many applications in physics and engineering, such as calculating torque, magnetic fields, and fluid dynamics. It is also used in computer graphics to calculate surface normals and in 3D modeling to determine the orientation of objects.

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