Determining the last column so that the resulting matrix is an orthogonal matrix 
Determine the last column so that the resulting matrix is an orthogonal
  matrix $$\begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} & ? \\ \dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{6}} & ? \\ 0 & \dfrac{2}{\sqrt{6}} & ? \end{bmatrix}$$

Can anyone please provide hints to solve this?
 A: Let $[ x \ y \ z]$ be the vector you are looking for, then it must satisfy
$$x \frac{1}{\sqrt{2}} +y \frac{1}{\sqrt{2}}+z (0)= 0$$
which is $y = - x$ and
$$x \frac{1}{\sqrt{6}} -y \frac{1}{\sqrt{6}}+z \frac{2}{\sqrt{6}} = 0$$
which is $z = \frac{1}{2}y - \frac{1}{2}x$. Using $y = -x$ in $z = \frac{1}{2}y - \frac{1}{2}x$, we get
$$z = -x$$
So the vector you are looking for has the form $[x, \ -x, \ -x]$
Choose $x = \frac{1}{\sqrt{3}}$ (for example), we get $[\frac{1}{\sqrt{3}}, \ -\frac{1}{\sqrt{3}}, \ -\frac{1}{\sqrt{3}}]$, which is a possible last column. 
A: Hint: remember that the cross product of two vectors of length 3 is orthogonal to both argument vectors and that an orthogonal matrix's columns are themselves an orthogonal set of vectors…
A: A matrix is orthogonal if all the column vectors are unit vectors and any two columns have dot product zero. Now write the unknown last column as $(x\ y\ z)^T$.  You will get two equations when when you insist it be orthogonal to the known first and second columns.
So solve a system of 2 equations an $x,y,z$. There will be infinitely many solutions. Now the condition on length will bring down the solutions to 2 (a vector and its negative).
A: Using SymPy:
>>> from sympy import *
>>> x, y, z = symbols('x y z')
>>> M = Matrix([[1/sqrt(2),  1/sqrt(6), x],
                [1/sqrt(2), -1/sqrt(6), y],
                [        0,  2/sqrt(6), z]])
>>> simplify(M.T * M)
Matrix([
[                1,                       0,       sqrt(2)*(x + y)/2],
[                0,                       1, sqrt(6)*(x - y + 2*z)/6],
[sqrt(2)*(x + y)/2, sqrt(6)*(x - y + 2*z)/6,      x**2 + y**2 + z**2]])

Thus, we obtain the following system of equations
$$\begin{aligned} x + y &= 0\\ x + z &= 0\\ x^2 + y^2 + z^2 &= 1\end{aligned}$$
which is easy to solve. Alternatively, we can take the $2$ known columns (which are orthonormal), form a $3 \times 2$ matrix, transpose it and then compute its null space:
>>> (Matrix([[1/sqrt(2),  1/sqrt(6)],
             [1/sqrt(2), -1/sqrt(6)],
             [        0,  2/sqrt(6)]])).T.nullspace()
[Matrix([
[-1],
[ 1],
[ 1]])]

All that is left to do is to normalize the vector that spans the null space. Switching its sign is legal.
