I'm trying to formulate an exercise related to linear algebra, and for that I need two vectors in $\mathbb{R^3}$ which
- have unit length,
- are orthogonal to one another,
- don't have any zero coordinates and
- can be represented using short decimal expansions of at most two digits
The key to answering this question lies in the consideration of Pythagorean quadruples. Enumerating (not only primitive) quadruples $(a,b,c,d)$ with $a^2+b^2+c^2=d^2$ I looked for those cases where $d=2^m5^n$ for some $m,n\in\{0,1,2\}$, i.e. those where $\frac1d$ has a finite decimal expansion with no more than two decimals. Then for each possible value $d$ I looked at all corresponding triples $(a,b,c)$, trying to find pairs which, for any permutation and any choice of signs, resulted in a zero dot product.
It turns out that, up to permutation and sign changes, there is exactly one such pair with two decimals:
$$ \left\langle\left(\begin{array}{r}0.36\\0.48\\0.80\end{array}\right), \left(\begin{array}{r}0.48\\0.64\\-0.60\end{array}\right)\right\rangle=0 $$
If you allow for one more decimal, you get 14 more combinations:
$$ \left\langle \left(\begin{array}{r} 0.024 \\ 0.640 \\ 0.768\end{array}\right), \left(\begin{array}{r} 0.768 \\ 0.480 \\-0.424\end{array}\right) \right\rangle=0 \qquad \left\langle \left(\begin{array}{r} 0.024 \\ 0.640 \\ 0.768\end{array}\right), \left(\begin{array}{r} 0.640 \\-0.600 \\ 0.480\end{array}\right) \right\rangle=0 \\ \left\langle \left(\begin{array}{r} 0.096 \\ 0.360 \\ 0.928\end{array}\right), \left(\begin{array}{r} 0.872 \\-0.480 \\ 0.096\end{array}\right) \right\rangle=0 \qquad \left\langle \left(\begin{array}{r} 0.096 \\ 0.360 \\ 0.928\end{array}\right), \left(\begin{array}{r} 0.480 \\ 0.800 \\-0.360\end{array}\right) \right\rangle=0 \\ \left\langle \left(\begin{array}{r} 0.096 \\ 0.480 \\ 0.872\end{array}\right), \left(\begin{array}{r} 0.360 \\ 0.800 \\-0.480\end{array}\right) \right\rangle=0 \qquad \left\langle \left(\begin{array}{r} 0.152 \\ 0.480 \\ 0.864\end{array}\right), \left(\begin{array}{r} 0.864 \\ 0.360 \\-0.352\end{array}\right) \right\rangle=0 \\ \left\langle \left(\begin{array}{r} 0.152 \\ 0.480 \\ 0.864\end{array}\right), \left(\begin{array}{r} 0.480 \\-0.800 \\ 0.360\end{array}\right) \right\rangle=0 \qquad \left\langle \left(\begin{array}{r} 0.168 \\ 0.224 \\ 0.960\end{array}\right), \left(\begin{array}{r} 0.576 \\ 0.768 \\-0.280\end{array}\right) \right\rangle=0 \\ \left\langle \left(\begin{array}{r} 0.168 \\ 0.576 \\ 0.800\end{array}\right), \left(\begin{array}{r} 0.224 \\ 0.768 \\-0.600\end{array}\right) \right\rangle=0 \qquad \left\langle \left(\begin{array}{r} 0.192 \\ 0.480 \\ 0.856\end{array}\right), \left(\begin{array}{r} 0.744 \\-0.640 \\ 0.192\end{array}\right) \right\rangle=0 \\ \left\langle \left(\begin{array}{r} 0.192 \\ 0.480 \\ 0.856\end{array}\right), \left(\begin{array}{r} 0.640 \\ 0.600 \\-0.480\end{array}\right) \right\rangle=0 \qquad \left\langle \left(\begin{array}{r} 0.192 \\ 0.640 \\ 0.744\end{array}\right), \left(\begin{array}{r} 0.480 \\ 0.600 \\-0.640\end{array}\right) \right\rangle=0 \\ \left\langle \left(\begin{array}{r} 0.352 \\ 0.360 \\ 0.864\end{array}\right), \left(\begin{array}{r} 0.360 \\ 0.800 \\-0.480\end{array}\right) \right\rangle=0 \qquad \left\langle \left(\begin{array}{r} 0.424 \\ 0.480 \\ 0.768\end{array}\right), \left(\begin{array}{r} 0.480 \\ 0.600 \\-0.640\end{array}\right) \right\rangle=0 $$
There are even some three-dimensional orthonormal bases among these. Again with no zero entries, and variations of those below might be obtained from sign changes, permutations of the rows or columns, and transposition.
$$ \left(\begin{array}{r} 0.024 & 0.768 & -0.640 \\ 0.640 & 0.480 & 0.600 \\ 0.768 & -0.424 & -0.480 \end{array}\right) \qquad \left(\begin{array}{r} 0.096 & 0.872 & 0.480 \\ 0.360 & -0.480 & 0.800 \\ 0.928 & 0.096 & -0.360 \end{array}\right) \\ \left(\begin{array}{r} 0.152 & 0.864 & -0.480 \\ 0.480 & 0.360 & 0.800 \\ 0.864 & -0.352 & -0.360 \end{array}\right) \qquad \left(\begin{array}{r} 0.192 & 0.744 & 0.640 \\ 0.480 & -0.640 & 0.600 \\ 0.856 & 0.192 & -0.480 \end{array}\right) $$
All of the above was found with quite a bit of brute force enumeration, so there might be more elegant approaches to finding these results. I also might very well have made a mistake.