# How to prove if 3 arbitrary vectors are coplanar

1. Dec 9, 2015

### cptcook

1. The problem statement, all variables and given/known data
Let s and t be scalars. show that u, v and su + tv are coplanar.
u and v are arbitrary vectors,

2. Relevant equations

3. 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.

2. Dec 9, 2015

### Staff: Mentor

Try using the triple scalar product and some vector algebra to reduce the expression.

3. Dec 9, 2015

### cptcook

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?

4. Dec 9, 2015

### Staff: Mentor

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.

5. Dec 9, 2015

### cptcook

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.

6. Dec 9, 2015

### Staff: Mentor

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.

7. Dec 9, 2015

### Staff: Mentor

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?

8. Dec 9, 2015

### cptcook

even if i dont 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).

9. Dec 9, 2015

### cptcook

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. Dec 9, 2015

### Staff: Mentor

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?
When you calculate $n \cdot (su + tv)$ all of the terms drop out, which is what you want to happen.

Last edited: Dec 10, 2015
11. Dec 11, 2015

### jk22

Maybe you know that uxv is ortogonal to u hence you could deduce u. (uxv) ?

12. Dec 11, 2015

### Staff: Mentor

I don't see how that gets the OP closer to showing that u, v and su + tv are coplanar.

13. Dec 12, 2015

### LCKurtz

@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. Dec 13, 2015

### jk22

This is a bit too easy : you could use the definition of the plane spanned by u and v