Triangle's altitude intersection point with its base The points $A(0, 0, 0)$, $B(−2, −1, 2)$, $C(0, 3, 4)$ are given. Find the equations of the line that contains altitude of the triangle ABC with initial point B.
Here's some more data if you want to approach finding point $D$. Or if you have other idea.
$\vec{AB} = [-2,-1,2] \quad \Longrightarrow ||\vec{AB}|| = 3$
$\vec{AC} = [0, 3, 4] \quad \quad \Longrightarrow ||\vec{AC}|| = 5$
$\vec{BC} = [2, 4, 2] \quad \quad \Longrightarrow ||\vec{BC}|| = 2\sqrt{6}$
$Area = 5\sqrt{2}$
$Area = \frac{1}{2} ||\vec{AC}||h \quad \Longrightarrow h = 2\sqrt{2} $
More or less this is the situation:

If I find point $D$ then I can find an equation of line passing through two points. That was my idea. No idea how to find point $D$ though. 
Unless there's simpler way to find the line, would someone explain how to find point $D$?
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Another idea is finding line containing points $A$, $C$ in its parametric form, and then the point $D$ would be described by that parametric form. Then, I can find vector $\vec{BD}$ and knowing that it has to satisfy the
$$\vec{BD} \circ \vec{AC} = 0$$
equation I can probably find the value of $t$ parameter. After that substitute that value into $D$ and find the line passing through $B$ and $D$.
This should work fine, however I am still glad to meet any alternatives.
 A: Another way to find $D$:
$\vec{AD}=\left(\dfrac{\vec{AB}\cdot\vec{AC}}{|\vec{AC}|^2}\right)\vec{AC}$
A: A vector parallel to the altitude of the triangle passing through $B$ will be perpendicular to $\overline{AC}$ and in the plane of $\triangle ABC$.
You can find a normal vector to the triangle by taking the cross product of any two of the vectors parallel to the sides. Then take the cross product of this normal vector with $\overline{AC}$ to get a vector parallel to the altitude (call it $\vec{A}$).
Take the coordinates of your point $B$ and add $\vec{A}$ to it, and you're done.
A: An alternative that uses homogeneous coordinates: The point $D$ is the intersection of the line through $A$ and $C$ and the perpendicular plane to it through $B$. The intersection of a plane $\mathbf\pi$ with the line through points $\mathbf p$ and $\mathbf q$ is $(\mathbf\pi^T\mathbf p)\mathbf q-(\mathbf\pi^T\mathbf q)\mathbf p$.  
The components of the homogeneous vector $\mathbf\pi$ that represents a plane are just the coefficients in the Cartesian equation $ax+by+cz+d=0$ of the plane. In this case, we have $$\mathbf\pi = (C-A;-(C-A)\cdot B) = (0,3,4,-5),$$ so the homogeneous coordinates of $D$ are given by $$\left((0,3,4,-5)\cdot(0,0,0,1)\right) (0,3,4,1)-((0,3,4,-5)\cdot(0,3,4,1)) (0,0,0,1).$$ Dehomogenize by dividing through by the last coordinate to obtain the inhomogeneous Cartesian coordinates of $D$.
