How to get remaining unit normals by purely geometric reasoning instead of calculating?

Question

Here "~" represent the vector.

1. Draw the tetrahedron AOBC through vertices A(1; 0; 0), the origin O(0; 0; 0), B(0; 2; 0) and C(0; 0; 3) in the standard cartesian 3D-frame.
2. Calculate the outward unit normal ~nABC on polygon ABC obligatory by using its two edges [AB] and [BC], respectively captured by their corresponding free vectors ~AB and BC. Simplify your result as far as possible by pen and paper.
3. Finally also give the outward unit normals ~nAOC, ~nAOB and ~nCOB on the remaining polygons by purely geometric reasoning instead of calculating.

There was no problems with a drawing and I got ~nABC (6/7; 3/7; 2/7) Cross product ~AB x ~BC (6; 3; 2) and it's length ||~AB x ~BC||=7. However, I have no idea how to get other normals by only geometric reasoning...

Answer 1

Simply note that the others three faces lie on $x-y$, $y-z$ and $z-x$ plane, therefore

• ~$nAOC=(0,-1,0)$
• ~$nAOB=(0,0,-1)$
• ~$nCOB=(-1,0,0)$

Note also that since we are looking for the outer normal we have

• ~$nABC=(-6/7,-3/7,-2/7)$

Answer 2

$OA$ is along $\hat x$, $OB$ is along $\hat y$, so the normal will be along $\hat z$, since $\hat z\perp \hat x$ and $\hat z\perp \hat y$. Since the tetrahedron is in the first octant, choose vectors pointing away from that.