How to calculate the height of a triangle without using vector cross product

Question

So I'm interested what would be the quickest and the most elementary of ways of solving the following problem:

Points $A(-6, -8, -1), B(4, 1, 1)$ and $C(2, -5, -4)$ are the vertices of a triangle. Calculate the shortest distance from vertex $C$ to the edge $AB$ of the triangle (that is, calculate the height from vertex $C$ to edge $AB$). You mustn't help yourself with vector cross product on your way to the solution. Good luck.

What is the quickest and easiest way to calculate that?

Very grateful for your time taken.

Answer 1

Let $a := |B-C|,\; b: = |A-C|,\; c := |A-B|\;$ be the lengths of the three sides of triangle $ABC$, and let $T$ be the area of the triangle which is $\sqrt{(-a+b+c)(a-b+c)(a+b-c)(a+b+c)}/4$ by Heron's formula. But $\;T = hc/2\;$ where $h$ is the height we seek, so solve it for $h$ for the answer.

For your example, as a check, $\; a \approx 8,\; b \approx 9,\; c \approx 13,\; h \approx 5.$

Answer 2

The line $AB$ is given by the parametric equation

$$AB \ldots\quad A + t(B-A) = (-6, -8, -1) + t(10, 9, 2) = (10t - 6, 9t - 8, 2t - 1)$$

for $t \in \mathbb{R}$.

Let $C_0$ be the point on $AB$ where the height intersects $AB$.

$C_0$ is attained for some $t_0 \in \mathbb{R}$

$$C_0 = (10t_0 - 6, 9t_0 - 8, 2t_0 - 1)$$

for which the vector $\vec{CC_0}$ is orthogonal to $(10, 9, 2)$.

We have:

\begin{align}\langle C - C_0, (10, 9, 2)\rangle &= \langle (2,−5,−4) - (10t_0 - 6, 9t_0 - 8, 2t_0 - 1), (10, 9, 2)\rangle\\ &= 101 - 185t_0 \end{align}

Therefore $t_0 = \frac{101}{185}$ so $C_0 = \left(-\frac{20}{37}, -\frac{571}{185}, \frac{17}{185}\right)$.

Now you can calculate the height as $d(C_0, C)$.