Cross Product of Two Vectors

[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