How do I prove that there are only two possible orthonormal basis? Let $x,y\in\mathbb{R}^3$ be unit vectors such that $\{x,y,(0,0,1)\}$ is an orthonormal subset of $\mathbb{R}^3$.
Let $z\in \mathbb{R}^3$ such that $\{x,y,z\}$ is an orthonormal subset of $\mathbb{R}^3$.
How do I prove that $z=\pm(0,0,1)$?
 A: You need to find a vector of norm $1$ such that it is orthogonal to vectors $x$ and $y$.
First method: write down the linear system of equations on coordinates of $z$ derived form orthogonality condition: the linear space of solutions has dimension $1$. After applying the normalisation you obtain two possible solutions. Then again, you already know that the solution $z=(0,0,1)$ fits, hence the conclusion.
Second method: use vector product. We know that the vector orthogonal both to $x$ and $y$ (if they are not colinear) has the form $\alpha x\times y$, with $\alpha\in \Bbb R$. The renormalisation gives you two pssible values of $\alpha$. After that we use the fact that $(0,0,1)=\pm x\times y$ and conclude.
A: The orthogonal space to $x,y$ have dimension 1. In your case the orthogonal is spanned by $(0,0,1)$. Since $z$ is orthogonal to $x$ and $y$; $z=\lambda(0,0,1)$ where $\lambda$ is a scalar. Since $z$ is normed you find that $\lambda=-1,1$.
A: Hints: Generally, if $(e_{i})_{i=1}^{n}$ is an (ordered) orthonormal basis of $\mathbf{R}^{n}$, an arbitrary vector $z$ is expressed as
$$
z = \langle z, e_{1}\rangle e_{1} + \dots + \langle z, e_{n}\rangle e_{n}
  = \sum_{i=1}^{n} \langle z, e_{i}\rangle e_{i}.
$$
By hypothesis, $\{x, y, (0, 0, 1)\}$ is orthonormal, and the unit vector $z$ is orthogonal to $x$ and to $y$.
