Finding conditions of $\vec w$ to be in a subspace

Question

Determine conditions of a,b,c so that the vector $\vec w=(a,b,c)$ is in span $S$ where $S=\{(2,1,0),(1,−1,2),(0,3,−4)\}$

Use this result to find a basis for span $S$.

Since those three vectors are linearly independent, the subspace spans all of $R^3$, so shouldn't the condition only be that a,b,c are real numbers?

Answer 1

For instante, $S$ is a linearly dependent set of vectors in $\mathbb R^3$, since $$(2,1,0)=2(1,-1,2)+(0,3,-4)$$ So, we can remove the vector $ (2,1,0) $ to obtain a linearly independent set. Thus $$\begin{align} \operatorname{span}(S) &= \operatorname{span}( \{(1,-1,2),(0,3,-4)\} ) \\ &= \{ t(1,-1,2)+s(0,3,-4) : \, t,s \in \mathbb R \} \\ &= \{ (t,-t+3s,2t-4s) : \, t,s \in \mathbb R \} \end{align}$$ Hence, the conditions of $a$, $b$ and $c$ so that $(a,b,c)\in \operatorname{span}(S)$ are : $$-2a+4b+3c=0$$ (that is, $(a,b,c)$ lies in the plane that generates $(1,-1,2)$ and $(0,3,-4)$).

Answer 2

Actually now I see that they are not linearly independent. I was thinking how since they are not direct scalar multiples of eachother, they are independent. But adding scalar multiples of two of those vectors does result in the third one.