How to find two vectors orthogonal to the gradient space of the feasible set? After finding the solution to the minimization of $f(x,y,z)=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2$ constrained by $h(x,y,z)=x^2+y^2+z^2-1=0$ by using the Lagrange multiplier method:
$$\nabla f(x,y,z) = 2 \begin{pmatrix} x-\frac{1}{2}\ \\ y-\frac{1}{2} \\ z-\frac{1}{2}\end{pmatrix}=\begin{pmatrix} 2x-1\ \\ 2y-1 \\ 2z-1\end{pmatrix}, \nabla h(x,y,z)=\begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$$
such that: $\begin{pmatrix} 2x-1\ \\ 2y-1 \\ 2z-1\end{pmatrix}=\lambda \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}, x=y=z=\frac{1}{2(1-\lambda)}$
solving for $\lambda$ using constraint: $3\left(\frac{1}{2(1-\lambda)}\right)^2=1 \Rightarrow 2(1-\lambda)= \pm \sqrt{3} \Rightarrow \lambda=1 \pm \frac{\sqrt{3}}{2}$
which yields: $x=y=z=\pm \frac{1}{\sqrt{3}}$
making the final solution: $(x,y,x,\lambda)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1 - \frac{\sqrt{3}}{2}\right)$
Using that solution, it was suggested to use Gramm-Schmidt orthogonalization to construct two independent vectors orthogonal to $\nabla h\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ which will be in the tangential space of the feasible set. 
I understand that I could just take two solutions to $a\frac{1}{\sqrt{3}}+b\frac{1}{\sqrt{3}}+c\frac{1}{\sqrt{3}}=0$ where $\vec{v}=(a,b,c)$ are the orthogonal vectors, but I'm curious how to implement Gramm-Schmidt with just one vector because as soon as you get to the second iteration, it seems that you need another vector. Is the idea to construct another linearly independent vector first and then use Gramm-Schmidt?

The optical inspection method works. The target vector is of the form
$$
u =
\left[ \begin{array}{c}
  1 \\ 1 \\ 1
\end{array} \right]
$$
Two orthogonal vectors are
$$
v_{1} =
\left[ \begin{array}{r}
  1 \\ 0 \\ -1
\end{array} \right], 
\quad
v_{2} =
\left[ \begin{array}{r}
  1 \\ -1 \\ 0
\end{array} \right].
$$
The process of Gram and Schmidt will orthogonalize the pair.
Update: once again @amd has a more elegant solution. Form the third vector as a cross product
$$
 v_{2} = u \times v_{1} 
$$

Final solution
$$
%
 u = \frac{1}{\sqrt{3}}
\left[ \begin{array}{c}
  1 \\ 1 \\ 1
\end{array} \right], \quad
%
 v_{1} = \frac{1}{\sqrt{2}}
\left[ \begin{array}{r}
  1 \\ 0 \\ -1
\end{array} \right], \quad
%
 v_{2} = \frac{1}{\sqrt{6}}
\left[ \begin{array}{r}
  1 \\ -2 \\ 1
\end{array} \right]
%
$$
