Given position vector of points A, B, find the equation of perpendicular bisector of AB in a vector form. I am trying to solve the problem: A, B are given point. a is the position vector of A, and b is the position vector of B. I want to find the equation of perpendicular bisector of AB in a vector form.
I solved this problem in a regular form, which I first set A $(x_1,y_1)$ B $(x_2,y_2)$, then the equation of this line is 
$$ (x_2-x_1)x+(y_2-y_1)y-\dfrac{x_2^2-x_1^2}{2}-\dfrac{y_2^2-y_1^2}{2}=0$$ 
But I don't know how to convert it in to something in the form of 
$$\vec{r}=\vec{r}_0+t\vec{v}=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle $$
Would someone give me a hint? Thanks in advance.
 A: It is likely more helpful to set out the problem using vector forms for the relevant lines.  The vector $ \ \vec{c} \ = \ \vec{AB} \ $ is  $ \ \vec{c} \ = \ \vec{b} \ - \ \vec{a} \ $ (since $ \ \vec{a} \ + \ \vec{c} \ = \ \vec{b} \ $ ) .  It appears that you are working in $ \ \mathbb{R}^2 \ $ , so we have 
$$ \vec{a} \ = \ \langle x_1 \ , \ y_1 \rangle \ \ , \ \ \vec{b} \ = \ \langle x_2 \ , \ y_2 \rangle \ \ , \ \ \vec{c} \ = \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
It is convenient to use $ \ \vec{a} \ $ as the "initial" vector with a parameterization that places $ \ t  = 0 \ $ there (point A) and places $ \ t = 1 \ $ at $ \ \vec{b} \ $ (point B) , with the direction given by $ \ \vec{c} \ $ .  The vector $ \ \vec{c} \ $ then lies along the line AB having parametric equation 
$$ \ \vec{r} \ = \ \vec{a} \ + \ t \ \vec{c} \ = \ \langle x_1 \ , \ y_1 \rangle \ + \ t \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
The midpoint is then located at  $ \ t = \frac{1}{2} \ $ , said point being at the head of position vector $ \ \langle  \frac{x_1 \ + \ x_2}{2} \ ,  $ $ \ \frac{y_1 \ + \ y_2}{2}  \rangle \   $ .
The perpendicular bisector of $ \ \overline{AB} \ $ can be defined by a vector which has a zero "dot product" (scalar product) with $ \ \vec{c} \ $ .  If we say that this line has a slope $ \ m \ $ , we can write its direction as $ \ \langle  1 \ , \ m  \rangle \ $ , and so
$$ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \cdot \ \langle  1 \ , \ m  \rangle \ \ = \ \ (x_2 - x_1 ) \ + \ m \ (y_2 - y_1) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ m \ \ = \ \ - \frac{x_2 - x_1}{y_2 - y_1} \ \ . $$
(Note that this is just what we expect from the "product of slopes of perpendicular lines" theorem, since the direction $ \ \vec{c} \ $ has slope $ \ \frac{y_2 - y_1}{x_2 - x_1} \ $ . )
Since $ \ y_2 - y_1 \ \neq \ 0 \ $ , we can then also write the direction for the perpendicular bisector as $ \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ $ .  (It is also acceptable to attach the "minus sign" to the first component instead -- thus, $ \ \langle y_1 - y_2 \ , \ x_2 - x_1 \rangle \ $ -- which is a vector pointing in the opposite direction on the perpendicular bisector line.
Finally, we can establish a parametrization of the perpendicular bisector with the midpoint of $ \ \overline{AB} \ $ at $ \ \tau = 0 \ $ , making it the "initial" vector for the bisector.  The vector form for this line may then be written as
$$ \vec{R} \ = \ \langle  \frac{x_1 \ + \ x_2}{2} \ ,  \ \frac{y_1 \ + \ y_2}{2}  \rangle \ + \ \tau \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ \ . $$
This method can be extended to $ \ \mathbb{R}^3 \ $ , except that the perpendicular bisector is ambiguous to the extent that it lies in a plane for which $ \ \vec{AB} \ $ is its normal.  To pin things down, we would need to specify a point in that plane that the bisector passes through.  (This is sometimes given as a textbook or exam problem.)  
A: Hint: The geometrical meaning of the formula $\vec r=\vec r_0+t\vec v$ is a line that passes in $\vec r_0$ (for $t=0$) and is directed along $\vec v$. So what you need is a point that the line goes through, and a direction. The point is easy - that's simply the midpoint between $\vec a$ and $\vec b$. 
The direction is a bit more subtle. Are you familiar with [cross products][1]?
[1]: http://en.wikipedia.org/wiki/Cross_product im just kidding this isn't right :'D I have no idea what im doing my bad person :'D
A: Note, for the question to be meaningful, we must be working in a plane (otherwise 'the' perpendicular bisector is not defined). Let $\hat{z}$ be a vector pointing out of the plane.
We want the equation of a line which passes through the midpoint of $\mathbf{a}$ and $\mathbf{b}$ and is perpendicular to $\mathbf{a}-\mathbf{b}$ and perpendicular to $\hat{z}$. Such a line is given by:
$$\frac12(\mathbf{a}+\mathbf{b})+\lambda (\mathbf{a}-\mathbf{b})\times \hat{z} $$
