1. Apr 10, 2010

### spoc21

1. The problem statement, all variables and given/known data

Prove that vectors $$\vec{u}$$, $$\vec{v}$$and $$\vec{w}$$ are coplanar if and only if vectors $$\vec{u}$$, $$\vec{v}$$and $$\vec{w}$$are linearly dependent

2. Relevant equations

3. The attempt at a solution

This may sound like an awkward question, but I am having much difficulty proving this. I tried to multiply each vector by a scalar, but got stuck, below is my solution, up until I got stuck.Any help is much appreciated!

We need to write $$\vec{u}$$= c1 $$\vec{v}$$+ c2 $$\vec{w}$$as:

[x1, y1, z1] = c1 [x2, y2, z2] + c2 [x3,y3, z3]

[x1, y1, z1] = [c1 x2, c1 y2, c1 z2] + [c2 x3, c2 y3, c2 z3]

x1 = c1 x2 + c2 x3
y1 = c1 y2 + c2 y3
z1 = c1 z2 + c2 z3

This is where I got stuck, and Im not sure what to do next.

Thanks,

2. Apr 10, 2010

### gabbagabbahey

Hmm.. I think the easiest way to prove this might be to realize that if 3 vectors are coplanar, then there will be some normal vector $\textbf{n}$ (the normal to the plane) that is perpendicular to all three and then look at some dot (and perhaps also cross) products.

3. Apr 10, 2010

### spoc21

Thats basically the problem, I'm supposed to prove this without using either dot or cross products Also, do you think that my working is fine, or will not work?

Thank you very much

4. Apr 10, 2010

### Staff: Mentor

This is an if and only if proof, which means you have to prove two things:
1. u, v, and w are coplanar ==> u, v, and w are linearly dependent.
2. u, v, and w are linearly dependent ==>u, v, and w are coplanar.

To get you started, if u, v, and w are coplanar, then u = c1v + c2 w for some nonzero constants c1 and c2. So what can you say about the equation u - c1v - c2 w = 0?

5. Apr 10, 2010

### spoc21

That means that they are linearly dependent...I completely understand the concept, but am stuck on the proof part; I'm a little confused on how to prove this using c1 and c2..

Thanks.

6. Apr 10, 2010

### gabbagabbahey

Not necessarily, if v and w are collinear, this need not be true. You need to examine two cases:

(1) At least two of the vectors are collinear
(2) None of the vectors are collinear.

7. Apr 10, 2010

### Staff: Mentor

I was afraid that might not be enough. It's a real restriction not having access to the dot product and cross product.

8. Apr 10, 2010

### spoc21

I think you're on the right track, could you please elaborate on your method..

Thanks!

9. Apr 11, 2010

### Staff: Mentor

Per gabbagabbahey's suggestion, look at two cases for u, v, and w.
1. At least two of the vectors are collinear.
2. None of the vectors is collinear.

In the first case, you can assume without loss of generality that u and v are collinear, i.e., that u = kv, for some constant k. Use that to show that the equation c1u + c2v + c3w = 0 has a solution where not all the constants are zero.

In the second case, use what I said toward the end of post 4 to show that the vectors are linearly dependent (i.e., that the equation c1u + c2v + c3w = 0 has a solution where not all the constants are zero).

10. Apr 12, 2010

### spoc21

ok, so I get it up till c1u + c2v + c3w = 0...How exactly do I show that c1u + c2v + c3w is equal to 0, for non zero scalars c1, and c1..

11. Apr 12, 2010

### Staff: Mentor

Depends on which case you're working on. If it's the first case, u = kv, so replace u in the c1u + c2v + c3w = 0 equation.

If it's the second case, see what I said in post 4.

12. Apr 12, 2010

### spoc21

it is the second case, where u = c1v + c2w, for non zero scalars, I don't understand how to actually prove what you said in post 4..or writing that would be sufficient..

Thanks,

13. Apr 12, 2010

### spoc21

ok, I get it...I was wondering how you got the u - c2v - c3w?

sorry, I just started vectors, and am a little confused with them..

14. Apr 12, 2010

### Staff: Mentor

If u = c1v + c2w, then u - c1v - c2w = 0.

So here I have a solution to the equation c1u + c2v + c3w = 0 where not all the constants are zero.

15. Apr 12, 2010

### spoc21

ok, so I get 0 when I substitute u = c1v + c2w into the equation: is this correct?, since these combinations need to be linearly dependent
inorder to be coplanar, which we proved, as u = c1v + c2w...

16. Apr 12, 2010

### Staff: Mentor

No, that's not it at all. Please tell me the definition of linear dependence, in the context of three vectors u, v, and w.

17. Apr 12, 2010

### spoc21

Linear dependence:

c1u + c2v + c3w= 0

18. Apr 12, 2010

### Staff: Mentor

Haven't you noticed that exactly the same equation is used to define linear independence? IOW, there is more to dependence or independence than just that equation. What does the complete definition say?

19. Apr 12, 2010

### spoc21

ok, so from what I've been taught, if you can express one of the vectors as a linear combination of the other vectors, then they are linearly dependent vectors...please remember that this is high school level math. I'm really confused on how to prove this, like I asked a tutor online , and he said that since these vectors are coplanar, so all we need to prove is that u - c1v + c2w = 0, so that c1, and c2 are non zero scalars...How could you prove this?
help is much appreciated..

Thanks!

20. Apr 12, 2010

### Staff: Mentor

This is not how linear dependence is defined.

For example, consider u = <1, 0, 0>, v = <0, 1, 0> and w = <1, 1, 0>.

I can set c1u + c2v + c3w = 0, and that alone doesn't tell me anything, because this equation is always true if c1 = c2 = c3 = 0. For these vectors, however, it turns out that 1u + 1v - 1w = 0 is also true, so I have found a solution for the equation c1u + c2v + c3w = 0, where not all of the constants are zero.

Here's another example, this time with u and v the same, but w = <1, 0, 1>.

I can again set c1u + c2v + c3w = 0, which I can always do, no matter what vectors I'm talking about. If I solve for the constants this time, I have
c1<1, 0, 0> + c2<0, 1, 0> + c3<1, 0, 1> = <0, 0, 0>
==> c1 + c3 = 0
c2 = 0
c2 + c3 = 0

Substituting 0 for c2 in the 3rd equation gives me c3 = 0. Substituting 0 for c3 in the first equation gives me c1 = 0.

The only solution to the equation c1u + c2v + c3w = 0 is c1 = c2 = c3 = 0, and there are no other solutions.

The upshot of this is that for an equation c1u + c2v + c3w = 0, u, v, and w are linearly independent if the only solution is c1 = c2 = c3 = 0, AKA the trivial solution. For the same equation, u, v, and w are linearly dependent if there is some solutions with at least one of the constants not equal to zero. The solution with all three constants equal to 0 applies in both cases.

Now, if you have u = av + bw, is there a solution to the equation u1 - av - bw = 0 other than the trivial solution? There obviously is.

21. Apr 14, 2010

### spoc21

Hi,

So this is what I am assuming (am very uncertain about it, and would appreciate it if you could rectify any mistakes):

lets assume that scalars: s and t, instead of c1, and c1 (to make it simple)

For these three vectors, u, v, andw to be coplanar:

u = sv + tw...(1)

For u, v, andw to be linearly dependent, we must have:

au + bv + cw (all scalars, not all of which are 0)

rearranging the first equation (1), we get:

u = sv + tw

u-sv-tw = 0...(2)

For constants we have 1(u), -s(v), and -t(w) = 0

Since we have one constant which is not 0, 1..we can assume that vectors u, v, and w are linearly dependent..

So now we know that u, v, and w are only coplanar, when vectors u, v, and w are linearly dependent..

Thanks!

22. Apr 14, 2010

### Staff: Mentor

This is mostly fine, but needs some tweaking.
This is the part most in need of change. The above should be the equation au + bv + cw = 0, and there must be one and only one solutions; namely a = b = c = 0.
Yes.
I don't know what you're trying to say here. Are you saying that 1u = 0? -sv = 0, -tw = 0? That's not the case at all.
No, we don't assume that - we know that because of the definition of linear dependence.
You have a lot of extra words that you don't need.

Starting with u = sv + tw...

u = sv + tw for some nonzero values of s and t
==> 1u - sv - tw = 0 has a non-trivial solution for the coefficients of the vectors; namely 1, -s, and -t
==> u, v, and w are linearly dependent.

23. Apr 14, 2010

### spoc21

I'm trying to say that since we have 1 as a constant, we know that it is nonzero, confirming that the vectors are linearly dependent (am not sure, but was told by another expert)

Also, I was wondering if you could please elaborate on what you mean, when you say that the equation has only one solution, specifically, a=b=c=0...Could please elaborate on your reasoning a bit further, as I'm a little confused right now....

Help is much appreciated.
Thanks!

24. Apr 15, 2010

### Staff: Mentor

OK, here's a simple example with the vectors u = <2, 1> and v = <-3, -1.5>. You should be able to realize that v = (-3/2)u or equivalently, that u = (-2/3)v.

v = (-3/2)u
<==> (3/2)u + 1v = 0
Since this is a linear combination of u and v that adds up to the zero vector and the coefficients of u and v are not all zero (in fact, neither one is zero), then by definition, u and v are linearly dependent. It's also worth noting that u and v are collinear.

Here's another simple example, this time with u = <2, 1> and v = <1, 2>. I set up the equation c1u + c2v = 0. Clearly c1 = c2 = 0 satisfies this equation, but I need to find out if there are other solutions (u and v lin. dependent) or this is the only solution (u and v lin. independent).

c1u + c2v = 0
<==> c1<2, 1> + c2<1, 2) = 0
<==> 2c1 + c2 = 0 and c1 + 2c2 = 0
By any of a number of methods, we can find that the only solution is c1 = 0, c2 = 0, so these vectors are linearly independent.

25. Apr 15, 2010

### spoc21

Ok, I understand..thank you...I attempted the solution, and have attached it in pdf format, and would really appreciate it if you please take a look, possibly commenting on it (its not very long , just one page)..

help is much appreciated.
Thanks a lot!

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