<s>[INST] 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?
 [/INST] 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. 
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