How to determine equation of tangent line in space? 

Let $c : [a,b] \longrightarrow \Bbb R^3$ be a differentiable function given by $c(t) = \left (x(t),y(t),z(t) \right ),$ $a \leq t \leq b$ which traces out a curve in the three dimensional space as $t$ varies from $a$ to $b.$ Let $\overrightarrow {r}(t)$ be the position vector of $c(t)$ with respect to the origin for $a \leq t \leq b.$ Let $t_0 \in [a,b].$ Let $l(t)$ be a parametrization of the tangent line to the curve $c$ at $c(t_0)$ for $c \leq t \leq d.$ Let $\overrightarrow {s} (t)$ be the position vector of the $l(t)$ with respect to the origin. Then the equation of the tangent line to the curve $c$ at $c(t_0)$ is given by $$\overrightarrow {s} (t) = \overrightarrow {r} (t_0) + (t-t_0)\ \overrightarrow {r'} (t_0)$$ where $\overrightarrow {r'} (t_0) = \left (x'(t_0),y'(t_0),z'(t_0) \right ).$


I know that the vector $\overrightarrow {r'} (t_0)$ determines the direction of the tangent line to the curve $c$ at $c(t_0)$ and owing to that reason the vector $\overrightarrow {r'} (t_0)$ is called the tangent vector to the curve $c$ at $c(t_0).$
Now lets move back where we studied straight lines in the three dimensional space in the context of solid geometry. We know that any straight line can be specified entirely by a point and a vector such that the line is passing through the given point and parallel to the given vector which is called the support of the line. The way we formulated the equation of a line through vector algebra in the following way.
Let $L$ be a line in the three dimensional space which passes through the point $M_0(x_0,y_0,z_0)$ and parallel to the vector $\overrightarrow a = l \hat i + m \hat j + n \hat k.$ Let $M(x,y,z)$ be any point on the line line except $M_0.$ Then we can easily see that the vector $\overrightarrow {MM_0} = (x-x_0) \hat i + (y-y_0) \hat j + (z-z_0) \hat k$ is collinear to the vector $\overrightarrow a.$ Therefore there exists non-zero scalar $s$ such that $\overrightarrow {MM_0} = s \overrightarrow a$ which simplfies three equations $$\begin{align*} x & = x_0 + s l\\ y & = y_0 + sm\\ z & = z_0 + sn \end{align*}$$ These three equations together represent the equation of the line $L$ in parametric form where $s$ is a parameter. As $s$ varies from $-\infty$ to $\infty$ the points $(x,y,z)$ trace out line $L.$
Now coming back to the case of tangent line we now have a point $c(t_0)$ through which the line tangent to the curve $c$ passes and we also have a direction vector namely $\overrightarrow {r'} (t_0).$ I know then by my concept of three dimensional solid geometry what I have just discussed above that the tangent line to the curve is completely specified by the point $c(t_0)$ and the direction vector $\overrightarrow {r'} (t_0).$ So the equation of the tangent line is given by $$\overrightarrow {s} (t) = \overrightarrow {r} (t_0) + \alpha \ \overrightarrow {r'} (t_0)$$ where $\alpha$ is a parameter just as before. How do I get back from here the equation of the tangent line in the prescribed format?
Any help or valuable suggestion regarding this will be highly appreciated.
Thank you very much for your valuable time for reading.
 A: This came up in another question not too long ago. The use of the word “the” is a poor choice here (as it is in the all too common usage “the eigenvector.”) There is obviously an infinite number of parameterizations of the tangent line. If all you’re interested in is the tangent line per se, there’s no particular reason to choose the parameterization in the quote block over another. You can certainly offset the parameter by some constant without changing the line that’s being described: replace $\alpha$ by $t-t_0$ to go from your parameterization to the other one. Why stop there, though? One might just as well normalize the tangent vector $\vec {r'}(t_0)$ for an arclength parameterization of the line—the parameter represents distance along it. If there’s any parameterization that deserves to be called “the” parameterization of a curve, one could argue that it’s that one.  
That said, the parameterization $\vec r(t_0)+(t-t_0)\vec {r'}(t_0)$ used in the quoted text has the virtue of being a linear approximation (in fact, the best one in a specific technical sense) to the curve at $r(t_0)$. It’s the equivalent in $\mathbb R^3$ of approximating a differentiable function $f:\mathbb R\to\mathbb R$ at $x_0$ as $f(x)\approx f(x_0)+(x-x_0)f'(x_0)$. This also explains why the same parameter is used for both the tangent line and the original curve, which otherwise would be a dubious practice. It would’ve been nice had the author explained this particular choice as “the” parameterization of the tangent line. Perhaps that’s done elsewhere in the text.
