How to determine if vectors span a plane?

Question

I have the set of three vectors $S=\{(1,2,-1),(2,0,1),(3,2,0)\}$.I observe that the last vector is linear combination of the first two vectors.Is it possible to say that the set $S$ spans a plane without reducing the matrix? Only observing the set can we conclude that the span of the vectors is a plane?

Answer 1

Because $(1,2,-1)+(2,0,1)=(3,2,0)$ then $$\text{span}(\{(1,2,-1),(2,0,1),(3,2,0)\})=\text{span}(\{(1,2,-1),(2,0,1)\})$$ and the last two vectors are linearly independent. The generated by two linearly independent vectors are, in fact, a plane.

Answer 2

If you write $S=\{v_1,v_2,v_3\}$ and you observe $v_3=v_1+v_2$, you can see that $\mathrm{span}(v_1,v_2,v_3)=\mathrm{span}(v_1,v_2)$, and because $v_1\neq0\neq v_2$ and $v_1\neq \lambda v_2$ for any $\lambda\in\mathbb{F}$, with $\mathbb{F}$ the field where the scalars from your vector space come from, you can see that $\{v_1,v_2\}$ is linearly independent and thus span a plane.

Explicitly row reducing the the matrix with the three vectors is not necessary if you provide a good (similar to mine) argument.

Answer 3

If any 2 vectors in S are not multiples of each other, then those 2 vectors span a plane. But that's is easier if you have a small set of vectors to check.

Answer 4

Let $\vec{u}=(1,2,-1)$, $\vec{v}=(2,0,1)$ and $\vec{w}=(3,2,0)$. We have $\mathrm{Span}\{\vec{u},\vec{v}\}=\mathrm{Span}\{\vec{u},\vec{v},\vec{w}\}$. The parametric equations of plane generated by $\vec{u}$ and $\vec{v}$ is $$ (x(t,r),y(t,r),z(t,r))=t\cdot \vec{u}+r\cdot \vec{v} $$