How to show that the span of vectors is a plane

Question

I am studying the course of affine and Euclidean goemetry. I am new to this course. I have studied two theorems of the affine combination.

1. The point $y$ is an affine combination of points $v_{1}, v_{2}, v_{3},...,v_{k}$ $iff$ $y-v_{1}$ is a linear combination of the points $v_{2}-v_{1}, v_{3}-v_{1}, ..., v_{k}-v_{1}.$
2. $y$ comes from affine combination of the points $v_{1}, v_{2}, v_{3},...,v_{k}$ $iff$ homogenuous form of $y$ comes in the linear combination of homogenuous form of vectors $v_{1}, v_{2}, v_{3},...,v_{k}.$

Now the question is if {${v_{1}, v_{2}, v_{3}}$} is basis for $R^{3}$, then how can I show that Span{${v_{2}-v_{1}, v_{3}-v_{1}}$} is plane in $R^{3}$ and Aff{${v_{1}, v_{2}, v_{3}}$} is plane through $v_{1}, v_{2}, v_{3}$. I don't know how to utilize these theorems to handle this problem.

Answer 1

Any 2-dimensional subspace is a plane. So if you show that v$_2$-v$_1$ and v$_3$-v$_1$ are linearly independent, then you'd be done.

I'm confused what Aff{$v_2-v_1,v_3-v_1$} is. Do you mean the affine hull? Because the affine hull of 2 points is the line segment connecting the 2.