$A(1,1,1)$, $B(2,1,2)$, $C(3,2,1)$ and $D(2,3,2)$ form a tetrahedron. If $ABC$ is the base, then what is the height?

Question

$A(1,1,1)$, $B(2,1,2)$, $C(3,2,1)$ and $D(2,3,2)$ form a tetrahedron. If $ABC$ is the base, then what is the height?

I found out of the equation of the plane containing A, B and C. It is $$-x + 2y +z -2 = 0$$

Then I found out the distance between the plane and $D(2,3,2)$ using the following: $$\begin{align} distance &= \frac{-(2) + 2(3) + (2) - 2}{\sqrt{(-1)^2 + (2)^2 + (1)^2}} \\ &= \frac{4}{\sqrt 6} \\ &= 2 \sqrt{\frac 2 3} \end{align}$$

This distance should be the height, because it is the perpendicular distance between $D$ and $ABC$.

But, my book says the answer is $$\frac 13 \sqrt{\frac 23}$$

Which answer is correct? Also, if I'm wrong, why?

Answer 1

Your answer is correct.

An alternative solution is to compute with vectors, i.e. to evaluate

$$h=\frac{\frac16|\vec {AC}\cdot(\vec{AB}\times\vec{AD})|}{\frac12|\vec{AB}\times\vec{AD}|}$$