Find an orthogonal basis for the subspace described by $x − y − z = 0$. Find an orthogonal basis for the subspace described by $x − y − z = 0$.
First, find two linearly independent vectors in the span of the subspace above.
$v_1=\begin{pmatrix} 1\\ 1\\0 \end{pmatrix}$ $v_2=\begin{pmatrix} 1\\ 0\\1 \end{pmatrix}$
Then find the projection of $v_2$ on to $v_1$. this equals $\frac{v_2⋅v_1}{v_1⋅v_1}v_1=\begin{pmatrix} 1/2\\ 1/2\\0 \end{pmatrix}$
Now I need to find the orthogonal component. $\begin{pmatrix} 1\\ 0\\1 \end{pmatrix}-\begin{pmatrix} 1/2\\ 1/2\\0 \end{pmatrix}=\begin{pmatrix} 1/2\\ -1/2\\1 \end{pmatrix}$
Therefore, the orthogonal basis for the subspace is $\begin{pmatrix} 1\\ 1\\0 \end{pmatrix},\begin{pmatrix} 1/2\\ -1/2\\1 \end{pmatrix}$
Questions: Did I do this correctly? Do my steps make sense? Alternative solutions?

Equation of the surface is $x-y-z=0$. 
We have two independent variable here let us take them $x,y$ so we determine $z$ from this. By this you got $v_1=\begin{pmatrix} 1\\ 1\\0 \end{pmatrix} $, Now take assume another vector which is orthogonal to it, let it be $\begin{pmatrix} x\\ y\\z \end{pmatrix}$ so $x.1+y.1+z.0=0$ so $x+y = 0$,(the dot product must be zero) . Also it has to satisfy $x-y-z=0$. So we get $z=-2y, x=-y$ . Here only one independent variable so it is a straight line.
So we can take this vector as $v_2=\begin{pmatrix} 1\\ -1\\2 \end{pmatrix} $
