Find height $DH$ of a tetrahedron $ABCD$ determined by vectors $\vec{AB}=(1,0,-1)$, $\vec{AC}=(0,1,1)$, $\vec{AD}=(1,2,0)$.

Question

We have to find the height $DH$ of the tetrahedron $ABCD$, and know that $\vec {AB}(1,0,-1), \vec{AC}(0,1,1), \vec{AD}(1,2,0)$.

I tried to calculate the mixed product and use the relation that says the volume of tetrahedron is $\tfrac16(\vec{AB}, \vec{AC}, \vec{AD})$. Finding the height from here gets complicated. I don't know how to find area of the base from here.

Answer 1

Given $3$ points in $\mathbb{R}^n$ for any $n \geq 2$, below are two methods for finding the area of a triangle whose vertices are those points. In both, if the origin is not already one of its vertices, it makes work much easier to first translate the triangle such that this is the case.

$\underline{\mathbf{Method \ \#1}}$:

Apply the formula $A = \displaystyle \frac{1}{2}ab\sin(\theta)$, which gives the area of the triangle having two side lengths $a$ and $b$ and angle $\theta$ between them (this data uniquely determines the triangle per $SAS$ congruence). First, get the two side lengths from the Pythagorean theorem. Now to get the angle, suppose the triangle has vectors $\mathbf{A}$ and $\mathbf{B}$ as sides. We know of the dot product of two vectors $\mathbf{v}$ and $\mathbf{w}$:

$$\mathbf{v} \cdot \mathbf{w} = ||\mathbf{v}|| ||\mathbf{w}|| \cos(\phi)$$

where $\phi$ is the angle between them. Therefore, letting $\mathbf{A'}$ and $\mathbf{B'}$ denote the normalizations of $\mathbf{A}$ and $\mathbf{B}$ respectively, $\theta = \arccos(\mathbf{A'} \cdot \mathbf{B'})$. Note that one would not even have to compute this inverse cosine, because $\sin(\arccos(x)) = \sqrt{1 - x^2}$.

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$\underline{\mathbf{Method \ \#2}}$:

A triangle is uniquely determined by the lengths of its sides per $SSS$ congruence. It follows that a triangle's area is determined by the lengths of its sides, and this is conveyed in Heron's formula:

Theorem: If a triangle has side lengths $a, b, c$, then its area is equal to: $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ Where $s$ denotes the triangle's semiperimeter, $\displaystyle \frac{a+b+c}{2}$.

You can find the area of the base in question using this formula, first finding the lengths needed using the Pythagorean theorem.