Is the average of cross product of two vectros equal to the cross product of averages of individual vectors?
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Yes, it is. The magnitude is the product of the magnitudes of the vectors multiplied by the sine of the angle between them. If you join the tips of the vectors you divide the parallelogram into two triangles. The area of a triangle ABC is (1/2) bc sin A, so the parallelogram is twice that.
From the details, it seems that you have no problem with the magnitude of the cross product being the area of a parallelogram, but don’t like an area to have a direction. This is purely a convention, but a useful one because not all two dimensional areas in three dimensions are equal. They have an orientation.
Another way to define orientation is via dyads. A dyad is a pair of vectors that define a plane.
As you need a convention to decide which of two opposite directions to take, a cross product is regarded as a pseudo-vector.
If you move to more than three dimensions, a cross product can no longer be represented nicely as a vector. The wedge product was introduced to allow for this. In terms of cartesian coordinates, the wedge product can be represented as an anti-symmetric matrix.
For the wedge product in three dimensions, the diagonal elements are zero, and three of the off diagonal elements are the components of the cross product vector, while the other three are components of its negative.
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Faizan Ahmed
Alexander Farrugia
Alexander Farrugia, frequent user of linear algebra.
Answered Jul 4, 2017
As I’m understanding your question, your problem is not that a×b is perpendicular to both a and b, but that the magnitude of a×b has to be that of the parallelogram whose non-parallel sides are oriented along a and b.
Honestly, this just happens to be the case.
Suppose I want to create an operation × that, when applied to two vectors a and b, produces a vector perpendicular to both according to the right hand rule. (Why the right hand rule? Because that’s how our coordinate system works.) I would first make sure that the following is true:
i×j=k;j×i=−k.
Reasonable, right? Well, you may ask me “why not 2k, or 12k? Why k?” My answer would be “that’s the simplest answer out of all you mentioned that is perpendicular to i and j. And I prefer simple answers.”
So, in a similar vein, let’s decide that
j×k=i;k×j=−i
and
k×i=j;i×k=−j.
This is what the right-hand rule actually means, by the way. It means that the cross product of two consecutive vectors from the list i,j,k,i is positive, while it is negative for two consecutive vectors taken from the reversal of that list.
Anyway, we need another important rule. What are we going to do when we multiply two parallel vectors? For example, what’s i×i? There are infinitely many vectors that are perpendicular to vector i, so which one are we going to choose? The answer is: the zero vector. If you think about it, the zero vector is actually perpendicular to every vector! So let us also decide that
i×i=0
j×j=0
k×k=0.
That’s it. We’re finished.
Q: “What? We’re not even close to being finished? What about the cross products of all the other vectors apart from i, j and k?”
Alex: “You mean vectors like 2i−3j+4k? Oh, just open up the brackets as usual.”
Q: “What do you mean?”
Alex: “I mean something like this:”
(2i−3j+4k)×(3i−2j+k)
=6(i×i)−4(i×j)+2(i×k)−9(j×i)+6(j×j)−3(j×k)+12(k×i)−8(k×j)+4(k×k)
=6(0)−4(k)+2(−j)−9(−k)+6(0)−3(i)+12(j)−8(−i)+4(0)
=−4k−2j+9k−3i+12j+8i
=−6i+10j+5k.
Alex: “See how easy that was?”
Q: “Okay, I get it now. But is this the same as the cross product I know about already?”
Alex: “Sure. And, in order to show you that this is exactly the same as the cross product you know about, I’ll prove that the magnitude of a×b produced by my cross product is equal to |a||b|sinθ.”
Q: “Uh, okay, go ahead then.”
Alex: “Brace yourself, this is gonna be a mouthful.”
We are going to prove that |a×b|2=|a|2|b|2sin2θ. We are allowed to do this because the magnitude of a vector is non-negative, so we do not have any problems with taking square roots.
First of all, suppose a=a1i+a2j+a3k and b=b1i+b2j+b3k. Then, using the same technique as above:
a×b=(a1i+a2j+a3k)×(b1i+b2j+b3k)
=(a2b3−a3b2)i+(a3b1−a1b3)j+(a1b2−a2b1)k.
Thus
|a×b|2=(a2b3−a3b2)2+(a3b1−a1b3)2+(a1b2−a2b1)2.(1)
On the other hand,
|a|2|b|2sin2θ=|a|2|b|2(1−cos2θ)=|a|2|b|2−|a|2|b|2cos2θ
|a|2|b|2sin2θ=(a⋅a)(b⋅b)−(a⋅b)2
=(a21+a22+a23)(b21+b22+b23)−(a1b1+a2b2+a3b3)2
=(a21b21+a21b22+a21b23+a22b21+a22b22+a22b23+a23b21+a23b22+a23b23)−(a21b21+a22b22+a23b23+2a1a2b1b2+2a1a3b1b3+2a2a3b2b3)
=a21b22+a21b23+a22b21+a22b23+a23b21+a23b22−2a1a2b1b2−2a1a3b1b3−2a2a3b2b3
=(a21b22−2a1b2a2b1+a22b21)+(a21b23−2a1b3a3b1+a23b21)+(a22b23−2a2b3a3b2+a23b22).
|a|2|b|2sin2θ=(a1b2−a2b1)2+(a1b3−a3b1)2+(a2b3−a3b2)2.(2)
Comparing (1) and (2), the result is established.
(Note that, as a corollary, we have also proved that |a×b|2=(a⋅a)(b⋅b)−(a⋅b)2, a result that is interesting in its own right.)