Answer: pretty sure they’re all true
Step-by-step explanation:
1. Should be obvious once you realize that if they all are in the same plane, one of them is a linear combination of the other two (think about a mini R^2 situation on the plane)
2. If x,y,z are linearly independent, they all are in the same plane, so z can be written in terms of x and y. (Above reasoning again.)
3. Should be intuitively clear, but rigorously just do the contrapositive. If v1,v2 are linearly dependent, av1 + bv2 = 0 for some a,b, so you can do av1 + bv2 + 0v3 = 0 which means v1,v2,v3 are dependent as well.
4. Intuitively they all are in the same plane. But if you want to be rigorous, you can choose, for example, 5v1 + 0v2 + 0v3 = 0, so they are linearly dependent.
5. What this is really saying is there is no scalar k such that v1 = kv2. It should be clear they are independent, but again if you want to be rigorous, suppose (for contradiction) there are a,b such that av1 + bv2 = 0. Then v1 + (b/a)v2 = 0, so v1 = -(b/a)v2. But this violates the “k” thing we said before! Hence a contradiction, so such a combination is not possible and thus they are linearly independent.
