Finding two vectors in a tetrahedron

Question

The question:

Consider the tetrahedron with vertices $(0, 0, 0)$, $(a, 0, 0)$, $(0, b, 0)$, and ($0, 0 ,c)$ and let $S$ be the side of the tetrahedron with vertices $(a, 0, 0)$, $(0, b, 0)$, and $(0, 0, c)$. By finding 2 vectors in $S$, find a unit normal to $S$.

I know how to find a unit normal given two vectors, but I am unsure how to find two vectors in this case.

Answer 1

Hint:

Let: $$ \vec A= (a,0,0)^T \qquad \vec B= (0,b,0)^T \qquad \vec C= (0,0,c)^T $$

take the vectors $\overrightarrow {AB}=\vec B-\vec A$ and $\overrightarrow {AC}=\vec C-\vec A$

Answer 2

There's an easier way to answer this question if you're not required to use the specified method.

The equation of the plane containing face $S$ is $$ \frac{1}{a} x + \frac{1}{b} y+ \frac{1}{c} z = 1 . $$ That means $$ \left( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \right) $$ is perpendicular to the plane. Now divide it by its length to normalize it.