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!
 A: I'm assuming you can find the equation of a line i) given two points and ii) given a point and slope and that you can do some other basic algebra which will show up. Comment if you don't understand anything.  


*

*Determine the equation of the line AC. In particular, we need its slope for the next step. 

*The slope of BD is the negative reciprocal of the slope of AC (so if for instance AC has slope equal to $\frac{2}{3}$ then BD has slope $-\frac{3}{2}$). This follows because BD is perpendicular to AC. 

*You know the slope of BD and one point (B) so you can find its equation. Do this. 

*Solve for D by finding the point of intersection of AC and BD. This can be done by equating their y-values. 

*Use the length formula $L=\sqrt{(\Delta x)^2 + (\Delta y)^2}$ as well as $\Delta y = m \Delta x$ (m is slope) to create a quadratic which you can solve for $\Delta x = x_t - x_d$. You can use this to find $x_t$.

*Solve for $y_t$ using the equation for BD. 
Comment if you get stuck following this. 
A: 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 $$T=D+n{\vec{BD}\over\|\vec{BD}\|}.$$
A: To find $DA$ it is enough to calculate the dot product of $BA$ and $CA$.
$$D-A = \frac{(C-A)}{|C-A|}\cdot\frac{\langle B-A,C-A\rangle}{|C-A|} = (C-A)\cdot\frac{\langle B-A,C-A\rangle}{\langle C-A, C-A\rangle}.$$
If you want a better intuition why this works, check out the Gram-Schmidt orthogonalization.
When you have $D = A + (D-A)$, you can find $T$ by extending $BD$ from $D$ for a desired length
$$T = D+n\cdot\frac{(D-B)}{|D-B|}.$$
I hope this helps $\ddot\smile$
