Vector Geometry find cross product.

Unart
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Homework Statement


Calculate the cross product of (3u+4w)xw assuming that
uxv=<1,1,0>, uxw=<0,3,1), vxw<2,-1,-1)

Homework Equations


Possible Relevant eqation:
i) wxv=-vxw
ii)vxv=0
iii)vxw=0 if and only w= λv for scalar λ or v=0
iV)(λv)xw=vx(λw)=λ(vxw)
V) (u+v)xw= uxw+vxw
ux(u+w)=uxv+uxw
vector-projection-formula.png


The Attempt at a Solution


So I have no real attempts, or attempts I feel were viable I feel like I'm doing guess work.
I've tried using equation letter V to some how try to single out one of the vectors using projection formula, but then I realized it was saying (v+w)xU instead of (vxw)xu. But as you can see that is down a wrong path.

I feel like this problem is an inside joke I'm just not getting lol.

How do you isolate each vector, they aren't orthogonal to each other. It's almost like the only thing these three vectors have in common are the parent vectors each play part in two of the product vectors. The also doesn't have ANYTHING like this, so it may be something really basic I'm forgetting as an example.
 
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Unart said:

Homework Statement


Calculate the cross product of (3u+4w)xw assuming that
uxv=<1,1,0>, uxw=<0,3,1), vxw<2,-1,-1)


Homework Equations


Possible Relevant eqation:
i) wxv=-vxw
ii)vxv=0
iii)vxw=0 if and only w= λv for scalar λ or v=0
iV)(λv)xw=vx(λw)=λ(vxw)
V) (u+v)xw= uxw+vxw
ux(u+w)=uxv+uxw
vector-projection-formula.png


The Attempt at a Solution


So I have no real attempts, or attempts I feel were viable I feel like I'm doing guess work.
I've tried using equation letter V to some how try to single out one of the vectors using projection formula, but then I realized it was saying (v+w)xU instead of (vxw)xu. But as you can see that is down a wrong path.

I feel like this problem is an inside joke I'm just not getting lol.

How do you isolate each vector, they aren't orthogonal to each other. It's almost like the only thing these three vectors have in common are the parent vectors each play part in two of the product vectors. The also doesn't have ANYTHING like this, so it may be something really basic I'm forgetting as an example.

Hint: What happens if, before you use any numbers, you apply the highlighted rule directly to your original problem?
 
Unart said:

Homework Statement


Calculate the cross product of (3u+4w)xw assuming that
uxv=<1,1,0>, uxw=<0,3,1), vxw<2,-1,-1)
Is there a typo here? The reason I ask is that there is no "v" in the cross product to be evaluated, ##(3\vec u + 4\vec w)\times \vec w##.
 
D H said:
Is there a typo here? The reason I ask is that there is no "v" in the cross product to be evaluated, ##(3\vec u + 4\vec w)\times \vec w##.

Not a typo, may be a typo on the book but I doubt it. However the book does equate Kilograms with a unit of force in one of it it's problems, so it wouldn't be a surprise. These parameters applies to six different cross-products, I understand how to do half of them. They are the following.

17. VxU which is just <-1,-1,0>, the negative of UxV
18. V x (u+v), or VxU + VxV, which equals to VxU = <-1,-1,0>
19. W x (U+V) which is just (UxW + VxW)*-1 (it's reversed) = <-2,-2,-2>
20. which I showed
21. (u-2v)x(u+2v); I don't know how to do.
22. (v+w)x(3u+2v); I also don't know how to do.

the professor asked for #20 and 21
 
Unart said:
21. (u-2v)x(u+2v); I don't know how to do.

Are you still talking about the vectors in the original post? What happens if you expand that out before you put in any numbers?
 
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