How to find three vector on $xy$-plane such that in $\Bbb R^3$ be perpendicular?

Question

How to find three vectors $v_1,v_2,v_3$ on $xy$-plan of same length in $\Bbb R^3$ such that after the following transformation be mutually perpendicular?

$$\phi:\Bbb R^2\times \{0\}\to \Bbb R^3,$$ $$v\mapsto v+e_z$$ where $e_z$ is $(0,0,1)$.

Answer 1

Identify $\mathbb{R}^2$ with $\mathbb{R}^2 \times \{0\} \subset \mathbb{R}^3$ throught the map $\mathbb{R}^2 \ni (x,y) \mapsto (x,y,0) \in \mathbb{R}^3$.
Let $v_1, v_2, v_3$ be the 3 vectors we seek and $\ell$ be their common length.

In order for $v_i + e_z$ to be orthogonal to $v_j + e_z$ for any distinct pair $\{ i, j \} \subset \{1,2,3\}$, we need

$$(v_i + e_z)\cdot(v_j + e_z) = 0 \iff v_i \cdot v_j = -1$$

This implies the angle between $v_i$ and $v_j$ equals to $\cos^{-1}\left(\frac{v_i \cdot v_j}{\ell^2}\right)$. Since this is the same for all 3 possible choices of $\{ i,j \}$, the angles among any pair of the 3 vectors are $\frac{2\pi}{3}$. This leads to

$$-1 = \ell^2 \cos\frac{2\pi}{3} = -\frac12\ell^2\quad\implies\quad \ell = \sqrt{2}$$

To construct an explicit solution, just pick any vector in $\mathbb{R}^2$ of length $\sqrt{2}$, rotate it clockwise and anticlockwise for $120^\circ$ and you are done. One obvious choice is pick $v_1$ to be one along $x$-axis and this leads to following solution:

$$v_1 = (\sqrt{2},0),\quad v_2 = \left(-\frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{\sqrt{2}}\right) \quad v_3 = \left(-\frac{1}{\sqrt{2}}, -\frac{\sqrt{3}}{\sqrt{2}}\right)$$

Answer 2

HINT:

The three final vectors must be perpendicular: $$ (v1+ez)\cdot(v2+ez)=0\\ (v2+ez)\cdot(v3+ez)=0\\ (v3+ez)\cdot(v1+ez)=0 $$ which is, noting that $v_i$ lie in the XY plane: $$ v1\cdot v2+1=0\\ v2\cdot v3+1=0\\ v3\cdot v1+1=0 $$ thus: $$ x1x2+y1y2+1=0\\ x2x3+y2y3+1=0\\ x3x1+y3y1+1=0 $$

Properly fixing 3 of the 6 coordinates, for example $v_1$ and $x_2$, makes the system with a unique solution.