vector calculus - line in vector form I'm really confused in the part of vector calculus by the book by Michael Corral.
In the part of the "Line through a point and parallel to a plane", parametric representation of a line L through a point P parallel to a vector v, has been described.

Theorem 1.16 given in the book(p.g. 31) states that it can be obtained by the formula:
r + tv, where t is the parameter,
The question down in page 32, example 1.19, tells us to write the line L, through the point P and parallel to the vector v, in the vector form. This is solutioned only by the above formula.  
But, as you can see in the diagram I have presentd above, which is much similar to the book, is the vector from the origin to Q. So, how is the vector even parallel, and even passing through the point P?
 A: One of the main confusions in writing a line in vector form is to determine what $\vec{r}(t)=\vec{r}+t\vec{v}$ actually is and how it describes a line.
For each $t_0$, $\vec{r}(t_0)$ is a vector starting at the origin whose endpoint is on the desired line.  You don't get a vector which points in the same direction as the line, but a vector ending on a line.  For different $t$-values, the endpoint of the vector $\vec{r}(t)$ gives different points along the line.
On the other hand, the vector $\vec{v}=\vec{PQ}$ is parallel to the given line (even though when this vector starts at the origin, it doesn't intersect the given line).
The geometric meaning of the equation $\vec{r}(t)=\vec{r}+t\vec{v}$ is as follows:  $\vec{r}$ is a vector from the origin to the line and $\vec{v}$ is a vector parallel to the desired line.  $\vec{r}$ is an arrow from the origin to a point on the line.  Then $t\vec{v}$ moves you from the endpoint of $\vec{r}$ in the direction of $\vec{v}$, which is parallel to the line.  Since we're starting on the line after moving $\vec{r}$, this means that we're adding a vector/segment along the line.  
I think of this as "giving directions", from the origin to get to a desired point on the line, walk to any point on the line (this is $\vec{r}$) and then walk in the direction of the line the desired distance (since $\vec{v}$ points in the direction of the line, $t\vec{v}$ scales the vector until it is the correct distance).  The result are directions from the origin to a point on the line.  Therefore, the resulting vector is not parallel to the line, but it starts at the origin and ends on the line.
A: The key to using vectors to represent a line not through the origin
is the correspondence between points and vectors described on
page 5 of your book.
Namely, since a vector can be described by three coordinates
$x,y,z$ and a point also can be described by three coordinates $x,y,z,$
we use the vector with coordinates $1,2,17$ (for example) as a
way to identify the point with coordinates $1,2,17.$
So when the book says that $\mathbf r + t \mathbf v$ is on the
line parallel to $\mathbf v$ through the point $P,$
it does not mean that the entire arrow labeled $\mathbf r + t \mathbf v$
in your figure is on that line.
It means only that when you draw the vector $\mathbf r + t \mathbf v$
so that its tail (starting point) is at the origin of your
coordinate system, the tip (ending point) of the vector is on
the desired line.
So another way to think about the vector formula $\mathbf r + t \mathbf v$
is that you take the three coordinates of the point $P$
and use them as the coordinates of a vector $\mathbf r.$
When you do this, the vector $\mathbf r$ gives the distance and
(if needed) the direction from the origin $(0,0,0)$ to the point $P.$
You take the vector $\mathbf v$ and multiply it by $t,$
that is, you multiply each of its coordinates by $t,$
to get a vector $t \mathbf v,$
and then you add the two vectors $\mathbf r$ and $t \mathbf v$
coordinate by coordinate to get the vector $\mathbf r + t \mathbf v.$
The vector $\mathbf r + t \mathbf v$ is described by three coordinates
which are also the three coordinates of a point on the desired line.
In short, don't look at the arrow, look only at the end point.

Aside:
I have a small quibble with the way this book describes vectors.
When the book defines vector, it says
a vector is a directed line segment.
This implies that a vector has six coordinates: the three coordinates
of its starting point and the three coordinates of its ending point.
And yet somehow the vector starting at $(1,1,2)$ and ending at $(4,5,7)$
is said to be "equal" to the vector starting at
$(0,0,0)$ and ending at $(3,4,5).$
I would rather say that a vector is only the displacement
(distance and direction)
from one point to another, which is in fact exactly the same for both
sets of points: $3$ units in the positive $x$ direction,
$4$ units in the positive $y$ direction, and $5$ units in the
positive $z$ direction will take you from $(1,1,2)$ to $(4,5,7)$
just as surely as it will take you from $(0,0,0)$ to $(3,4,5).$
According to my preferred definition, the arrow labeled
$\mathbf r + t \mathbf v$ in your figure
is not "the" vector $\mathbf r + t \mathbf v$;
it is not even a vector equal to $\mathbf r + t \mathbf v.$
Instead, it shows how the vector $\mathbf r + t \mathbf v$
relates the origin of coordinates to a point on the desired line:
if you start at the origin, and use the vector $\mathbf r + t \mathbf v$
to determine how far to travel in what direction,
you will get to a point on the line.
A: The vector equation is:
$$
\vec x=\vec p +t \vec v \qquad t \in \mathbb{R}
$$
that is:
$$
\begin{pmatrix}
x\\y\\z
\end{pmatrix}
=
\begin{pmatrix}
x_0\\y_0\\z_0
\end{pmatrix}+t
\begin{pmatrix}
v_x\\v_y\\v_z
\end{pmatrix}
$$
and, for $t=0$, ve have $\vec x=(x_0,y_0,z_0)^T=\vec p$. For other values of $t$ we have the other points on the line.
