Finding a vector parametric equation given P and Q equations? Find a vector parametric equation $r⃗(t)$ for the line through the points $P=(3,5,4)$ and $Q=(1,4,7)$ for each of the given conditions on the parameter $t$. 
If $r⃗ (0)=(3,5,4)$ and $r⃗ (7)=(1,4,7)$, then $r⃗ (t)= $?
I found the vector for $PQ$ and got $(-2,-1,3)$.
I then took the original $P$ and made $r(t) = (3,5,4) + t (-2,-1,3)$.
Not sure if I am just not really understanding the idea, but does anyone know how to solve this?
 A: The idea here is that $\vec{r}(t)$ will reach $P$ at $t=0$ and $Q$ at $t=7$. Your parametric equation of $\vec{r}(t)$ is good, but what you need to do next is scale the velocity vector $\langle-2,1,3\rangle$ by a factor $x$ such that $\vec{r}(t)=P$ for $t=0$ and $Q$ for $t=7$. Following this logic, we have
\begin{align*}
(3,5,4)&=(3,5,4)+0x\langle -2, -1, 3\rangle \\
(1,4,7)&=(3,5,4)+7x\langle-2,-1,3\rangle 
\end{align*}
which can be satisfied by $\displaystyle x=\frac{1}{7}$. So what is the intuition behind what we just did? You acquired $\langle -2, -1 3 \rangle$ by taking the difference of $P$ and $Q$, but we wanted $\vec{r}(t)$ to reach $Q$ at $t=7$ and not $t=1$, so we include the scale factor to compensate.
A: I know a lot of people who are confused by what a vector equation of a line mean. Even before you sit down to find the vector equation of a line, you must ask what does the equation really mean? It means that given the equation of the line you can tell just by manipulating a parameter the points through which the lines passes.
Now how do we uniquely define this equation? There are various ways to do this(vector and cartesian forms). In your problem they have asked the vector form. To write the vector equation we need two things.


*

*Scale

*Direction


The direction is given by the fact that the line passes through two points  $P=(3,5,4)$ and $Q=(1,4,7)$.We can now determine the direction by simply subtracting the x,y,z ($(2,1,-3)$)coordinates(the reverse is also possible). Clearly any point on the line will have this direction. But to arrive to that point we have to pass through $(3,5,4)$ as we start from the origin. The vector equation would be 
$$r(t)= (3,5,4) + (-2,-1,3)$$
But we have not taken care of the scale. Hence we need to multiply a constant term($\mathbf t$)to the direction. Hence the final equation is
$$r(t)=(3,5,4) + t(-2,-1,3)$$
A: This question will form two different answers in both conditions of $r(0)= (3,5,4)$ and when $r(7)=(1,4,7)$
First, start out by finding the Vector that passes through the points $p=(3,5,4)$ and $Q=(1,4,7)$. This will result in the vector $(v)=<-2,-1,3>$ just by subtracting the two points. 
With the first condition where $r(0)=P=(3,5,4)$. 
$r(t)=R_o+t_v$
where $R_o=(X_o,Y_o,Z_o)$.
$(3,5,4)=X_o+3t,Y_o+5t,Z_o+4t$
$X_o+3t=3,\; Y_o+5t=5$ and $Z_o+4t=4$. When $t=0$,
$X_o=3-3(0)=3$.  $Y_o=5-5(0)=5$ and $Z_o=4-4(0)=4$ therefore, $R_o= <3,5,4>$.
Hence for when $r(0)=P$, the vector equation is $r(t)= <3,5,4>+t<-2,-1,3>$. 
The same procedure matches for the condition when $r(7)=(1,4,7)=Q$
$r(t)=<X_o,Y_o,Z_o> +t<-2,-1,3>$.
therefore, $X_o-2t=1$, $Y_o-t=4$ and $Z_o+3t=7$. when $t=7$, 
$X_o=1+2(7)=15$, $Y_o=4+(7)= 11$ and $Z_o=7-3(7)= -14$. 
Hence, $r(t)= <15,11,14>+t<-2,-1,3>$. 
