So, the task goes like this: $A(0,1,0), B(3,1,2), C(2,0,1)$ are cordinates of a triangle $ABC$. Point $D$ is a height drawn from point $A$ , and point $E$ is a middle of side $BC$. Find cordinates of vector $ \overrightarrow{DE} $
I can find cordinates of $ \overrightarrow{BC} $, so we get $\overrightarrow{BC}(-1,-1,-1)$ and cordinate of point $E$ should be $E(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2})$
My idea is next: to make a equation of a slope on $BC$ side and after that height from point $A$ should cut that slope and create point $D$ there.
Equation of a slope should be : $\frac{x-3}{-1} = \frac{y-1}{-1} = \frac{z-2}{-1} $
Am I doing this right? If I am what should I do next?
Is it possible to do this without using a slope? I mean if I calculate $DC$ so I should get $DE= DC-EC$. My guess here should be to check angle of point $C$ and I should find $|DE|$, but how can I determinante components of this vector?
EDIT:
If $E$ is a midpoint of $BC$ then $$E = {1\over 2}(B+C) = ({5\over 2},{1\over 2},{3\over 2})$$
If $D(x,y,z)$ on $BC$ then $$ D = tB+(1-t)C = (3t,t,2t)+(2-2t,0,1-t)= (2+t,t,1+t)$$
and since $AD\bot BC$ $$(x,y-1,z)\cdot (1,1,1)=0\implies x+y+z =1$$
so $$ 2+t+t+1+t=1\implies t=-2/3\implies D({4\over 3},-{2\over 3},{1\over 3})$$