Why Are the Magnitudes of Vector Operations Surprisingly Equal?

[Mosaness]

1. See attached image please!
2. For part (a), I applied the cross product and got (-6i - 2k) for (\vec{A}x\vec{B}. I got (6i + 2k) for (\vec{B} x \vec{A}).

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

For part (c), the magnitude of \vec{C} was simply 12.65 and for the magnitude of two times (\vec{A} x \vec{B}) 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 (\vec{A}x\vec{B} is equal in magnitude but opposite in direction to (\vec{B} x \vec{A}), therefore, the magnitude for 2 times (\vec{A} x \vec{B}) ought to equal the magnitude of (\vec{C})

 

[ehild]

1. See attached image please!
2. For part (a), I applied the cross product and got (-6i - 2k) for (\vec{A}x\vec{B}. I got (6i + 2k) for (\vec{B} x \vec{A}).

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

For part (c), the magnitude of \vec{C} was simply 12.65 and for the magnitude of two times (\vec{A} x \vec{B}) 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 (\vec{A}x\vec{B} is equal in magnitude but opposite in direction to (\vec{B} x \vec{A}), therefore, the magnitude for 2 times (\vec{A} x \vec{B}) ought to equal the magnitude of (\vec{C}) (

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

 

[Mosaness]

I was doing part (d.) and the unit vector for \vec{C} was (-0.949i - 0.316k) and the unit vector for (\vec{A} x \vec{B}) was also (-0.949i - 0.316k). Therefore, the unit vector for \vec{C} is not twice as long as the unit vector for (\vec{A} x \vec{B}). 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 (\vec{A} x \vec{B}), as was the magnitude. And for (\vec{A} x \vec{B}), 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 (\vec{A} x \vec{B}) been half that of vector C, THEN the unit vector for (\vec{A} x \vec{B}) would have been half that of vector C.

 

[ehild]

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

 

[Muphrid]

Uh, I think you're overthinking this. What is the magnitude of any unit vector?

 

[Mosaness]

Well the magnitude will always be one. I WAS over thinking it! Oops