How do I meaningfully divide by a vector? 
How long does it take a baseball with velocity $(30, 20, 25) m/s$ to
  travel from location $r_1 = (3, 7,−9) m$ to location $r_2 = (18, 17, 3.5)m$?

I am thinking that it should be the displacement vector divided by velocity. but velocity is a vector and my text is adamant on not putting a vector in the denominator. I tried it anyway just to see what happens and I get  $(.5, .5,.5)s$ .The units do cancel out to give me seconds. The answer is listed in the book as .5 seconds. But how do I arrive there without breaking any rules? (I am assuming delta time is final time - 0.) 
 A: If you're standing at the origin, then the vector from you to the starting point of the ball is $\vec{r_1}$ and the vector from you to the end point is $\vec{r_2}$.

You need to find the vector $\vec{s}$ such that $\vec{r_1} + \vec{s} = \vec{r_2}$. The simplest way to do this is to note that when you give the vector as its three components, $x$, $y$ and $z$ then $r_{1x} + s_x = r_{2x}$ and likewise for the $y$ and $z$ components. Some quick arithmatic gives:
$$ \vec{s} = \left( 15, 10, 12.5 \right) $$
Now, you're told that this is a multiple of the velocity, $\vec{v}$, so:
$$ \vec{s} = n \vec{v} = \left( nv_x, nv_y, nv_z \right) $$
where $n$ is the number of seconds. Taking the $x$ component we get $s_x = 15 = nv_x = n 30$, so $n$ is 0.5. For completeness you need to check you get the same value of $n$ for the $y$ and $z$ components.
A: You take the magnitude of component of velocity in the direction of displacement.
And divide magnitude of displacement by it.
If $\vec{r}$ is your displacement and $\vec{v}$ is your velocity, the time taken would be
$$t = \frac{r^2}{\vec{v}.\vec{r}}$$, where $r$ is the magnitude of $\vec{r}$.
A: If at time $t=0$ the baseball is at location $\vec{r}_1$ and moving with speed $\vec{v}_1$ then withough any external influences its position as a function of time is
$$ \vec{r}(t) = \vec{r}_1 + \vec{v}_1 \; t =\vec{r}_2 $$
The above can be made into a scalar equation considering the direction of travel $\vec{e} = \frac{\vec{v}_1}{|\vec{v}_1|}$ with a dot product on both sides:
$$ (\vec{e}\cdot \vec{r}_2) =  (\vec{e}\cdot \vec{r}_1) + (\vec{e} \cdot \vec{v}_1)\, t $$
 $$ \vec{e}\cdot (\vec{r}_2-\vec{r}_1) = (\vec{e} \cdot \vec{v}_1)\, t $$
or 
$$ t = \frac{\vec{e}\cdot (\vec{r}_2-\vec{r}_1)}{\vec{e} \cdot \vec{v}_1} $$
This would produce the same result as
$$ t = \frac{\vec{v}_1\cdot (\vec{r}_2-\vec{r}_1)}{\vec{v}_1 \cdot \vec{v}_1} $$
since $|\vec{v}_1|$ would cancel from both the numerator as the denominator of the fraction.
If gravity is acting, then things get a little more complicated with 
$$ \vec{r}(t) = \vec{r}_1 + \vec{v}_1 \; t + \frac{1}{2} \vec{g}\, t^2 =\vec{r}_2 $$
requiring you to solve this by project the above along a vector $\vec{e}$ of your choosing. For example to find the time to hit the ground use $\vec{e}=\frac{\vec{g}}{|\vec{g}|}$.
