Finding a point on the line perpendicular to a line from another point I'm sorry if this is kind of basic, it's been a while since I took geometry. I did find this answer, but it requires 4 points and I only have 3.
I have three points $A$, $B$ and $C$ that form a non-right triangle. I know the Cartesian coordinates of these three points. Suppose I draw a perpendicular from $B$ through $AC$ where $D$ is the point where the perpendicular intersects $AC$. How can I find the Cartesian coordinates of a point $T$ along $\overrightarrow {BD}$ that is $n$ beyond $D$?
I drew a diagram to demonstrate what I'm trying to find.
Diagram

Any help would be appreciated, thanks!

You can solve this fairly easily with a few vector operations. 
Finding point $D$ comes down to finding the perpendicular projection of the vector $\vec{AB}$ onto $\vec{AC}$. That’s given by $$\vec{AB}_\parallel={\vec{AB}\cdot\vec{AC}\over\|\vec{AC}\|^2}\vec{AC}$$ and so $D = A+\vec{AB}_\parallel$.  
Now, recall that a line can be described parametrically using a point on the line and a direction vector. Since we’re measuring distances from $D$, we’ll use $D$ as our point and $\vec{BD}=D-B$ for the direction. Also, ince we want to move a specific distance along this line from $D$, we’ll normalize the direction vector by dividing by its length so that the resulting direction vector has unit length. That way, moving $n$ units along the line is simply a matter of multiplying the direction vector by $n$. Putting this together, we get $$T=D+n{\vec{BD}\over\|\vec{BD}\|}.$$
