How to prove if 3 arbitrary vectors are coplanar

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To prove that the vectors u, v, and su + tv are coplanar, the triple scalar product must equal zero. The discussion emphasizes that su + tv is a linear combination of u and v, indicating it lies in the plane formed by these vectors. Participants suggest using the cross product n = u × v and demonstrating that n · (su + tv) = 0 to establish coplanarity. There is some confusion about handling arbitrary vectors, but it is clarified that calculating the cross product and applying vector identities will lead to the desired result. Ultimately, the approach focuses on showing that the linear combination does not extend beyond the plane defined by u and v.
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Homework Statement


Let s and t be scalars. show that u, v and su + tv are coplanar.
u and v are arbitrary vectors,

Homework Equations

The Attempt at a Solution


Well, i suppose if they are coplanar, triple scalar product of them should be 0, right? and su+tv is evidently a linear combination of u and v. but i got a little confused when vectors are arbitrary and not really sure how to show it properly. any help would be appreciated, thanks.
 
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Try using the triple scalar product and some vector algebra to reduce the expression.
 
jedishrfu said:
Try using the triple scalar product and some vector algebra to reduce the expression.
ok i already went down that road and got lost, if i had numbers, it would be easier but how can i prove that they are coplanar if all i have is arbitrary vectors?
 
cptcook said:

Homework Statement


Let s and t be scalars. show that u, v and su + tv are coplanar.
u and v are arbitrary vectors,

Homework Equations

The Attempt at a Solution


Well, i suppose if they are coplanar, triple scalar product of them should be 0, right? and su+tv is evidently a linear combination of u and v. but i got a little confused when vectors are arbitrary and not really sure how to show it properly. any help would be appreciated, thanks.
Here's how I would do it:
1. Calculate n = u X v. (I'm assuming that u and v are vectors in R3)
2. Show that ##n \cdot (su + tv) = 0##.

Geometrically, for all real values of s and t, su + tv defines the plane that u and v lie in.
 
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Mark44 said:
Here's how I would do it:
1. Calculate n = u X v. (I'm assuming that u and v are vectors in R3)
2. Show that ##n \cdot (su + tv) = 0##.

Geometrically, for all real values of s and t, su + tv defines the plane that u and v lie in.

That is the general idea, yes but still, how to calculate n=u X v when i do not even have numbers, i think I'm missing your point here, can you be just a little more specific? i tried defining the vectors as u=(u1,u2,u3) and v=(v1,v2,v3) and then calculating the triple scalar product. it got me nothing.
 
cptcook said:
That is the general idea, yes but still, how to calculate n=u X v when i do not even have numbers, i think I'm missing your point here, can you be just a little more specific? i tried defining the vectors as u=(u1,u2,u3) and v=(v1,v2,v3) and then calculating the triple scalar product. it got me nothing.
Using the coordinates of u and v, as you show above, what do you get for u x v? You know how to calculate a cross product, right?

Once you get that, show that ##n \cdot (su + tv) = 0##. I don't think you need to use a scalar triple product.
 
U x V . ( sU + tV) = U x V . sU + U x V . tV

Using some more vector identities you can demonstrate that the expression evaluates to zero hence sU + tV is what?
 
Mark44 said:
Using the coordinates of u and v, as you show above, what do you get for u x v? You know how to calculate a cross product, right?

Once you get that, show that ##n \cdot (su + tv) = 0##. I don't think you need to use a scalar triple product.

even if i don't use triple scalar product and do what you suggest, i'd still have to calculate cross product (which i know how to do it), correct me if I'm wrong but;

u X v= det |(i,j,k),(u1,u2,u3),(v1,v2,v3)| right? that gives you a gigantic equation of nothing when calculated, which cannot be simplified or arranged, let alone taking the product of it with another arbitrary (su+tv).
 
jedishrfu said:
U x V . ( sU + tV) = U x V . sU + U x V . tV

Using some more vector identities you can demonstrate that the expression evaluates to zero hence sU + tV is what?

again, i cannot calculate the cross product of two arbitrary vectors in r3. it becomes a giant vector equation which is non-arrangeable. cross product seems like a dead end.
 
  • #10
cptcook said:
even if i don't use triple scalar product and do what you suggest, i'd still have to calculate cross product (which i know how to do it), correct me if I'm wrong but;

u X v= det |(i,j,k),(u1,u2,u3),(v1,v2,v3)| right? that gives you a gigantic equation of nothing when calculated, which cannot be simplified or arranged, let alone taking the product of it with another arbitrary (su+tv).
It gives you a vector expression with six terms, so it's a gross exaggeration to call it a "gigantic equation." Instead of dreaming up reasons why you can't do this, why don't you put some effort into actually trying to work the problem?
cptcook said:
again, i cannot calculate the cross product of two arbitrary vectors in r3. it becomes a giant vector equation which is non-arrangeable. cross product seems like a dead end.
When you calculate ##n \cdot (su + tv)## all of the terms drop out, which is what you want to happen.
 
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  • #11
Maybe you know that uxv is ortogonal to u hence you could deduce u. (uxv) ?
 
  • #12
jk22 said:
Maybe you know that uxv is ortogonal to u hence you could deduce u. (uxv) ?
I don't see how that gets the OP closer to showing that u, v and su + tv are coplanar.
 
  • #13
@cptcook: You don't need to mess with components. The above posts have shown the way. You yourself noted you just need to show the triple scalar product ##u\times v \cdot(su+tv)=0##. Post #11 reminds you that ##u \times v## is perpendicular to both ##u## and ##v##. Post # 7 shows you how to expand ##u\times v \cdot(su+tv)##. You should be able to see why that is zero without expanding by components.
 
  • #14
This is a bit too easy : you could use the definition of the plane spanned by u and v
 
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