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Cross Product of Two Vectors

  • Thread starter Mosaness
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  • #1
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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
ehild
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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
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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
ehild
Homework Helper
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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
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Uh, I think you're overthinking this. What is the magnitude of any unit vector?
 
  • #6
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Well the magnitude will always be one. I WAS over thinking it!!! Oops
 

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