Parametric equation of line? I have an assignment I'm doing where I am supposed to determine a parametric equation of a line orthogonal to a segment , let's say OP, and passes through the midpoint of this segment. 
What I have come up with so far is that since the line is orthogonal to OP, would it be easiest if I made up a plane M, and OP belongs to m as a line segment, then the line I'm searching for would be the plane's normal, since it is orthogonal to the plane, like the line should be to the segment OP?
The line passes through the midpoint of the segment. Would this be (x-x'/2, y-y'/2, z-z'/2), that would then be (2,2,2). I have also computed the directed segment of OP = P-O = (2,0,-2). 
I know I can compute the crossproduct of two linearly independent vectors, which would be perpendicular to both of them, but now I only have a segment in the plane, if I even should have the segment in a plane, since this is my own invention(the plane) But it just feels like I could compute the normal for it somehow, but I would need another vector or point right? So, this is where I'm stuck. Can someone set me going with this information that I got and that I have computed. How do I come up with a parametric equation for the line which passes midpoint of OP and is perpendicular to OP??
O = (1,2,3) 
P = (3,2,1)
 A: The parametric equation of a line looks like $$\begin{matrix}\vec r(t) = \vec r_0 + t\vec v, & t\in \Bbb R\end{matrix}$$ where $\vec r_0$ is a vector pointing from the origin to a point $P(x_1, x_2, \dots, x_n)$ on the the line and $\vec v$ is a vector parallel to the line.  Then $t$ is just your variable.
Notice that this looks very similar to the scalar equation of a line $y=mx+b$.

In your equation you're given a line segment defined by its two endpoints.  Let's say those endpoints are $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$.  Then first let's figure out what its midpoint is.  We need this because, as a point on the line we're trying to parametrize, it'll give us a point we can use to define $\vec r_0$.  The midpoint of the points $A(x_1,y_1,z_1)$ and $B(x_2, y_2, z_2)$ is the point $$M\left(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2},\frac{z_1 + z_2}{2}\right)$$  So our vector $\vec r_0$ is just $\vec{OM}$.
The next bit of info we need from that line segment is to know a vector parallel to it.  The two easiest vectors will be $\vec{AB} = \vec{OB} - \vec{OA}$ and $\vec{BA} = \vec{OA} - \vec{OB}$.  Compute one of those vectors.
Why do we need to know a vector parallel to the line segment when we want a line perpendicular to the line segment?  Because I said so :P.  Also because once we find that vector, say it's $\vec{AB} = a_1\hat e_1 + a_2\hat e_2 + a_3\hat e_3$ -- where $\{\hat e_1, \hat e_2, \hat e_3\}$ is an orthonormal basis for $\Bbb R^3$, we can use the definition of orthgonality to figure out a vector perpendicular to it.
The definition of orthogonality (perpendicularity) is $\vec a \text{ is orthogonal to } \vec b \iff \vec a \cdot \vec b = 0$.  Therefore, we can find $\vec v$ by simply using this formula with the vector $\vec{AB}$ that we already found.  So find any nonzero solution to the equation $\vec v \cdot \vec{AB} = 0$ and you'll have your $\vec v$.
Then, having $\vec r_0$ and $\vec v$, just write down your parametric equation $$\begin{matrix}\vec r(t) = \vec r_0 + t\vec v, & t\in \Bbb R\end{matrix}$$

Let's look at an example: say we're given the line segment whose endpoints are $A(1,1,1)$ and $B(2,0,2)$.
Then using the above we see that $$\operatorname{Midpoint}(A,B) = M\left(\frac{1+2}{2},\frac{1+0}{2},\frac{1+2}{2}\right) = M\left(\frac 32, \frac 12, \frac 32\right) \\ \implies \vec r_0 = \vec{OM} = \frac 32e_1 + \frac 12e_2 + \frac 32e_3 \\ \vec{AB} = \vec{OB} - \vec{OA} = (2\hat e_1 +0\hat e_2 + 2\hat e_3) - (1\hat e_1 + 1\hat e_2 + 1\hat e_3) = \hat e_1 -\hat e_2+\hat e_3 \\ \implies \vec v \cdot \vec{AB} = (v_1\hat e_1 + v_2\hat e_2 + v_3\hat e_3) \cdot (\hat e_1 - \hat e_2 + \hat e_3) = 0 \\ \implies v_1 -v_2 + v_3 = 0$$  We only need $1$ solution (out of an infinite number of solutions) to the above equation, so at this point, we'll let $v_1$ and $v_2$ be easy numbers (that are not both $0$), like $v_1 = 1$ and $v_2=0$, and just solve for $v_3$. Solving, we can clearly see that $v_3=-1$ when $v_1 =1$ and $v_2=0$.  Therefore $$\vec v = 1\hat e_1 + 0\hat e_2 + (-1)\hat e_3 = \hat e_1 -\hat e_3$$  Putting all of this together, the equation of one of the lines perpendicular to the line segment $\overline{AB}$, passing through its midpoint is $$\require{enclose}\enclose{box}{\begin{matrix}\vec r(t) = \left(\frac 32e_1 + \frac 12e_2 + \frac 32e_3\right) + t(\hat e_1 -\hat e_3), & t\in \Bbb R\end{matrix}}$$
