Find the values of $a, b, c, d$ that make a set of vectors an orthogonal set I'm being asked to complete the last two vectors to make $\vec{x}, \vec{y}, \vec{z}$ an orthogonal set: 
$$\vec{x}=(1, 1, 1), \vec{y}=(a, b, 1), \vec{z}=(c, d, 1)$$
I thought I needed to use Graham-Schmidt, but it seems to lead me through a rabbit hole of arbitrary calculations that look wrong. Any help with this would be super appreciated. 
 A: In our case, orthogonality is translated to the following system of equations:
$$(1,1,1) \cdot (a,b,1) = a + b + 1 = 0\implies a=-(1+b)$$
$$(1,1,1) \cdot (c,d,1) = c + d + 1 = 0\implies c=-(1+d)$$
$$(a,b,1) \cdot (c,d,1) = ac+bd + 1 =0\implies b+d+2bd+2=0$$
Your problem is not asking you to find all possible values, but only one value to make the set orthogonal. To find one possible value, 
assume $b=-d$ and you will get:
$$-2b^2+2=0 \implies b^2=1 \implies b=\pm 1$$
So, you find two solutions for $(a,b,c,d)$: $\{(-2,1,0,-1), (0,-1,-2,1)\}$
Picking the first one, gives us that $\{(1,1,1),(-2,1,1),(0,-1,1)\}$ is an orthogonal set and solves your problem.
However, you could go further than this. Suppose that $b = \lambda d$ where $\lambda$ is a variable to obtain:
$$2\lambda d^2 + (\lambda + 1)d + 2 = 0$$
Therefore,
$$d = -\frac{\lambda + 1 \pm\sqrt{\lambda^2 - 14\lambda + 1}}{4 \lambda}$$
which is valid when $\lambda^2 - 14 \lambda + 1 > 0$
Now you can calculate $b$ and then $a,c$ respectively. Therefore, all possible $4$-tuples $a,b,c,d$ satisfying your conditions are parametrized by $\lambda$. 
A: Just pick $a$ and $b$ such that $a+b+1=0$. Then vectors $x$ and $y$ are
orthogonal. Consider their cross product; unless its last entry is zero
one can divide by it and get $z$ orthogonal to $x$ and $y$ with last entry $1$.
