Equation of height of a tetrahedron

Question

$A(2,3,-2)$, $B(3,0,-5)$, $C(1,2,4)$ and $D(-1,1,6)$ form a tetrahedron. How do I find the equation of the height $DH$? In previous points of the same problem I've found the equation of the base plane

$$ABC: 21x + 3y + 4z - 43 = 0$$

but I don't know if there is a way to use it here. I can't think of a way to find coordinates of $H$ either.

Answer 1

Hint:

$\vec{HD}$ must be parallel to $\begin{bmatrix} 21 \\ 3 \\ 4 \end{bmatrix}$

Furthermore, $\vec{HD}$ must passed through $D$.

Hence for some $k \in \mathbb{R}$,

$$\vec{OH}-\vec{OD}=k\begin{bmatrix} 21 \\ 3 \\ 4 \end{bmatrix}$$

$$\vec{OH}=\vec{OD}+k\begin{bmatrix} 21 \\ 3 \\ 4 \end{bmatrix}$$

Using the info that $H$ lies on the plane $ABC$, you should be able to solve for $k$ and hence know the coordinate of $H$.

Answer 2

The distance from a point $D$ to a plane $\pi$ of type $h(x,y,z)=ax+by+cz-k=0$ is found by inserting the coordinates of $D$ into $h$ and dividing by the squareroot of $a^2+b^2+c^2.$ This may end up of either sign, which has to do with which "side" of the plane $D$ lies on. This is the perpendicular distance from $D$ to $\pi,$ which it seems is what you want for the "height".

Answer 3

Let $a$, $b$, and $c$ be the vector from $A$ to $B$, $C$, and $D$, resp. Then the volume $V$ is the absolute value of $\frac16\det(a,b,c)$. On the other hand the area $F$ of triangle $ABC$ is $F=\frac{1}{2}\sqrt{\langle a,a\rangle\langle b,b\rangle-\langle a,b\rangle^2}$.

Now we may calculate $h$ from $V=\frac13F\cdot h$.