Find the parametric equation of the line. Find the parametric equation of the line from $ \ (2,1,-3) \ $ to $ \ (1,-2,4) \ $ for $ \ t \in [2,4] \ $.
Answer:
The parametric equation of the line will be 
$ \ l: \ t \left\langle 1,-2,4 \right\rangle+(2-t) \left\langle 2,1,-3 \right\rangle\ $
i.e., $ \ l: \ \left\langle t+4,-3t+2,7t-6 \right\rangle \ $
Am I right ?
i need confirmation of my work.
 A: You want to start with $(2,1,-3)$ at $t=2$ and get to $(1,-2,4)$ at $t=4$
$$x=2-(1/2)(t-2)$$
$$y=1-(3/2)(t-2)$$
$$z= -3 +(7/2)(t-2)$$
is the parametric equation for the line segment. 
A: I would write $t((1,-2,4)-(2,1,-3))+(2,1,-3)$ to parametrize the line... so $(-t+2,-3t+1,7t-3)$...is the line.  Then let $t$ range from $2$ to $4$...
You should be using: $t(1,-2,4)+(\color{blue}1-t)(2,1,-3)$
What you've got there is the line through $(4,2,-6)$ and parallel to the line you seek...
Now, what I did takes you from $(2,1,-3)$ at $t=0$ to $(1,-2,4)$ at $t=1$.  To adjust this, the transformation $t\rightarrow \frac t2-1$ does the trick.   That is, we get:  $(\frac t2+3,\frac{-3t}2+4,\frac{7t}2-10)$...
There was a little confusion about what you meant by "for $t\in [2,4]$", so I gave two interpretations...
A: Given a pair of points $P$ and $Q$, $R(t) = (1-t)P+tQ$, or, if you prefer, $P+t(Q-P)$, is a parameterization of the line segment joining them, with $t\in[0,1]$, $R(0)=P$ and $R(1)=Q$. If you want the parameter to range over some other interval $[a,b]$, you must scale and translate it by making the substitution $t \to (t-a)/(b-a)$, giving $$P+{t-a \over b-a}(Q-P)$$ or the more symmetric $${b-t \over b-a}P+{t-a \over b-a}Q.$$
