Given a triangle with points in $\mathbb{R}^3$, find the coordinates of a point perpendicular to a side Consider the triangle ABC in $\mathbb{R}^3$ formed by the point $A(3,2,1)$, $B(4,4,2)$, $C(6,1,0)$.
Find the coordinates of the point $D$ on $BC$ such that $AD$ is perpendicular to $BC$.
I believe this uses projections, but I can't seem to get started. I tried the projection of $AC$ onto $BD$ and $AB$ onto $BC$, but to no avail.
Any help is loved! Thanks.
 A: Here is one way, presumably unsuitable. 
The direction vector of $BC$ is $(2,-3,-2)$. 
A generic point $D_t$ on $BC$ is given by $(4,4,2)+t(2,-3,-2)$. 
The direction vector of $AD_t$ is $(1,2,1)+t(2,-3,-2)$. The dot product of this with $(2,-3,-2)$ is $-6+17t$. This dot product must be $0$. 
We end up with $D=\left(\frac{80}{17},\frac{50}{17},\frac{22}{17}  \right)$.
A: Edited:
There are two ideas used is this.
(1),$AD$ must be perpendicular to $BC$, so their dot product must equal 0.
(2),$D$ is on $BC$, so we must find an eqn. for line $BC$.
Let's start with (2).  We know that the direction vector of the line $BC$ is parallel to the vector $\overrightarrow BC$ which is $<2,-3,-2>$ and the line passes through $B(4,4,2)$. Therefore the parametric eqn. for line $BC$ must be $$x = 2t+4$$ $$y = -3t+4$$ $$z=-2t+2$$
This means that all points on the line $BC$, in particular, $D$, will have coordinates of the form $$D(2t+4,-3t+4,-2t+2)$$
Now let's move on to (1).
The vectors $\overrightarrow {AD}$ and $\overrightarrow{BC}$ are perpendicular to each other, so their dot product must be 0, i.e $$<3-(2t+4),2-(-3t+4),1-(-2t+2)> \cdot <2,-3,-2> = 0$$
All you have to do is to find $t$ and plug it back in.
