This question is from the 1st chapter of Intro to Linear Algebra, 4th Ed, by Gilbert Strang.
So please omit concepts which succeed this question: matrices, rank, REF, nullspace, $Ax=b$, linear independence, span, basis, dimension, dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear maps.
The linear combinations of $\mathbf{v} = (a,b)$ and $\mathbf{w} = (c,d)$ fill the plane unless $\color{green}{\text{complete this blank}}$.
Answer: … $\color{green}{\Large{[}} \mathbf{v}$ and $\mathbf{w}$ lie on the same line through $(0,0) \color{green}{\Large{]}}$.
$\Large{1.}$ How and why isn't the answer:
"$\mathbf{v}$ and $\mathbf{w}$ lie on A/ANY same line" ?
Why must it be the collinear one "through $(0,0)$" ?
$\Large{2.}$ How can this be generalised? Any catholic lessons from this question?
My weak (cp the PDF) definition of "unless" predicates upon this MSE Question
and P113 as printed on the page or P23 of 50 in the PDF Viewer:
http://courses.umass.edu/phil110-gmh/text/c04_3-99.pdf
$ A \text{ unless } B \; \equiv \; \color{ #B53389}{\text{Unless }} B, A \; \equiv \; \color{ #B53389}{\text{If not }} B, A \; \equiv \; \neg B \longrightarrow A $.
Best Answer
The linear combinations of the vectors $v,w$ can be denoted $span(v,w)=kv+cw$, where $v,w \in V$ and $k,c \in \mathbb{F}$ (scalars).
If the two vectors are linearly independent, informally consider the follow scenario:
Let $V = \mathbb{R^2}$. Let two vectors $v,w \in V$. Set $v = (x,0)$ and $w = (0,y)$. There is no scalar $k \in \mathbb{F}$, that when multiplied by the vector $v = (x,0)$, yields a vector of the form $w = (0,y)$. The two independent vectors, when combined linearly as $kv+cv$, make up a whole plane or $\mathbb{R^2}$. You can think of these vectors as the $x-axis$ and $y-axis$ coming together to form the $xy-plane$. When taking the set of all linear combinations of the vectors $v$ and $w$ we can obtain any other vector of the form $(x,y)$, and thus $span(v,w) = \mathbb{R^2}$.
Formally, the only combination $kv+cw$ that yields the $0$ vector is by setting each $k,v \in \mathbb{F}$ equal to $0$. Is is clear that the only scalars that satisfy the linear combination $k(x,0) + c(0,y) = (0,0)$ for the $0$ vector are $k = c = 0$, which implies that the two vectors in $\mathbb{R^2}$ are independent and span the plane.
Now, consider a separate case, where the vectors $v,w \in V$ are $(x,0)$ and $(2y,0)$, respectively. Let $span(v,w) = a(x,0) + b(2y,0)$ which denotes all possible linear combinations of the vectors. However, it is apparent that the $0$ can be obtained by more than just the "obvious" way which means setting $a = b = 0$ to obtain $(0,0) + (0,0) = (0,0)$. Setting $y = \frac{-1}{2}x$, and we get $a(x,0)+b(-x,0)$ which equals $0$ for $a = b = 0$ or $a = b = 1$, which shows that the two vectors are dependent and as a result do not span $\mathbb{R^2}$, so their linear combinations do not fill a plane.
It's not that $v$ and $w$ can't lie on the same line. It's just that if we have two vectors are on the same line, then they are two linearly dependent vectors in $\mathbb{R^2}$, and as a result, do not fill a plane.