create orthogonal matrix from given rows

Question

Let's say I have the following matrix: \begin{bmatrix} \frac{1}{3} & \frac{2}{3} & 0 & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} & \frac{2}{3} & 0 \\ a & b & c & d \\ e & f & g & h \end{bmatrix}

How do I find the last $2$ rows such that this matrix is orthogonal? I know that it's orthogonal if $M*M^T = Id$, but this isn't helping me, nor does the gram-schmidt process.

Answer 1

I suggest to get inspiration from the first two rows. Two elements of each row cancel out each other, so it could the case for the rows $3$ and $4$. Start by setting $b=e=0$.

P.S.

The final answer would be $$ { (a,b,c,d)=(-\frac{2}{3},0,\frac{2}{3},\frac{1}{3}) \\ (e,f,g,h)=(0,\frac{2}{3},\frac{1}{3},-\frac{2}{3}) } $$

Answer 2

First of all, note that there is more than on answer.

Let$$W=\left\{(x,y,z,t)\in\Bbb R^4\,\middle|\,(x,y,z,t).\left(\frac13,\frac23,0,\frac23\right)=(x,y,z,t).\left(\frac23,-\frac13,\frac23,0\right)=0\right\}.$$Then a basis of $W$ will be $\{(-4,2,5,0),(-2,-4,0,5)\}$. It turns out that they're orthogonal. So, an orthonormal basis of $W$ is$$\left\{\frac1{3\sqrt5}\left(-4,2,5,0\right),\frac1{3\sqrt5}\left(-2,-4,0,5\right)\right\}.$$So, take $(a,b,c,d)=\left(-\frac4{3\sqrt5},\frac2{3\sqrt5},\frac5{3\sqrt5},0\right)$ and $(e,f,g,h)=\left(-\frac2{3\sqrt5},-\frac4{3\sqrt5},0,\frac5{3\sqrt5}\right)$. The determinant of the matrix will either be $1$ or $-1$. If it's $1$, you're done. Otherwise, exchange those two vecotrs.