In the text of Linear Algebra, by Steven_Levandosky, it is given for span of two vectors that:
$span(v_1, v_2)$ contains every line which passes through the line spanned by $v_2$ and is parallel to line spanned by $v_1$
The image of the page is here (which I am linking as do not know how to exactly reproduce the image given by a tool known to me. Request a tool to help in drawing such images).
The confusion part is in the statement given above (in the second line) , which is part of para. below (& also given on the image link):
Geometrically, consider some particular values of $c_1, c_2$. First fix $c_2=0$ and let $c_1$ vary all over of $\mathbb{R}$ to see that span contains the line spanned by $v_1$. Next, fix $c_2=1$ and let $c_1$ vary over $\mathbb{R}$ to see that $span(v_1, v_2)$ contains the line which passes through the head of $v_2$ and is parallel to the line spanned by $v_1$. By continuing to fix $c_2$ and let $c_1$ vary over $R$ we see that $span(v_1, v_2)$ contains every line which passes through the line spanned by $v_2$ and is parallel to the line spanned by $v_1$
My understanding of span is that it covers all possible (real) linear combinations of two vectors in a $2$-D plane, and three vectors in a $3$-D plane. So, it automatically means that in a $2$-D plane, that any third vector would be a linear combination of the other two linearly independent vectors ( by which I mean that they are not a multiple of each other). Similarly, for $3$-D plane, where need only $3$ linearly independent vectors to define any new vector in the plane.
The last line of the book's statement (in bold ) confuses me, as can only interpret it as:
every vector which passes through the vector spanned by all possible multiples of $v_2$ (i.e. found by multiplying by a scalar value of a real number, including $0$) and is parallel to any multiple of the vector $v_1$ by a real scalar $c_1$.
--
But, this interpretation does not explain to me how a third vector in the plane would be expressible as a linear combination of the two. For example, let the two linearly independent vectors be: $u = [1\,\, 2], v= [3 \,\, 2]$, and introducing a new vector, let $w = [3\,\, -2]$ means that $w$ is a linear combination of the other two, given by $w = 2v - 3u$. In fact, as $w$ is not a multiple of any of the two vectors $u, v$; so can take any two vectors of the three (i.e. $u,v,w$), and can express the third in terms of the other two; say $u = \frac23 v - \frac13 u$.
I cannot use this particular example to justify the confusing part (in bold) of the book.
Maybe this will help. Let's show in $\mathbb{R}^2$ that the following are equivalent:
To prove this, first suppose $z=cu+dv$ for some $c,d\in\mathbb{R}$. Now, consider the line $\ell=\{tu+dv:t\in\mathbb{R}\}$. $\ell$ is parallel to the line spanned by $u$ (which is the line $\{tu:t\in\mathbb{R}\}$) and crosses the line spanned by $v$ (namely, it contains the point $dv$ which is on the line $\{tv:t\in\mathbb{R}\}$). Furthermore, $z\in\ell$ since $z=cu+dv\in\ell$.
For the reverse direction, suppose that $\ell$ is a line parallel to the line spanned by $u$ and crosses the line spanned by $v$ and $z\in\ell$. Since $\ell$ is parallel to the line spanned by $u$, we have $\ell=\{tu+x_0:t\in\mathbb{R}\}$ for some $x_0\in\mathbb{R}^2$. Since $\ell$ crosses the line spanned by $v$, this means we can find some $t_0,t_1\in\mathbb{R}$ for which $t_0u+x_0=t_1v$. Finally, $z\in \ell$, so there is some $t_2\in\mathbb{R}$ with $z=t_2u+x_0$. Putting these together, we have $$ z=t_2u+x_0=(t_2-t_0)u+(t_0u+x_0)=(t_2-t_0)u+t_1v, $$ so $z$ is a linear combination of $u$ and $v$.