Planes in linear algebra I have a question, say I have two linearly independent vectors, then would there be only one plane in R3 containing these two vectors?
 A: It depends what you mean by "plane". If you mean "2-dimensional subspace" then yes, by definition. (If you're interested in what happens when you generalize to hyperplanes and hence any proper subspaces then it depends:


*

*In $\mathbb{R}^2$, the result is no.

*In $\mathbb{R}^3$, the answer is yes.

*In $\mathbb{R}^4$, and higher, the answer is no.
Proof of 1: Calling the vectors $v, w$, we have $\mathbb{R}^2 = span(v,w)$, so no proper subspace contains the vectors.
Sketch of proof of 2: Let two proper subspaces contain $v, w$ and show they are equal. This is not hard to do, let me know if you need advice on carrying out the proof. It starts by pointing out that the subspace has to be of dimension 2 to be both proper and contain $v,w$.
Sketch of proof of 3: There are infinitely many vectors $z$ such that $\{v,w,z\}$ is linearly independent, and this is a proper subspace in every case.)
But, to reiterate: Since "plane" means for you a 2-dimensional subspace, then there is only one "plane" containing a given set of 2 linearly independent vectors, and this is because the plane containing 2 linearly independent vectors would be in this case defined as their span, i.e. the set of all their linear combinations, which is unique.
