Finding general solution based on system of equation My question is from a homework set I found: In $\mathbb{R}^4$, and given 3 orthonormal vectors $u_1, u_2, u_3$, can we find a fourth vector $u_4$ such that all 4 vectors are orthonormal?
Given are $u_1 = [\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}], u_2 = [\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}], u_3 = [\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}]$
Let $u_4 = [a, b, c, d]$. Then the dot product of $u_4$ with any of the other vectors needs to be $0$. So we can encode this in the following system of equations:
$$
\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}d = 0 \\
\frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}d = 0 \\ 
\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c - \frac{1}{2}d = 0 
$$
Row-reducing this in matrix form gave me the following solutions:
$$
a+b+c+d = 0 \\
b - c = 0
$$
Now how to derive a general solution from here?
In the homework set they mention that the general solution would be $u_4 = [t, -t, -t, t]$ and I agree, but based on the simplified system of equations I wouldn't have been able to see that (expect that $b=c=t$, that's clear.) I think I'm missing some intermediate step to understand the given solution.

The system of equations you got, is correct
$\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}d = 0 \\
\frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}d = 0 \\ 
\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c - \frac{1}{2}d = 0$
From here these equations are reduced (you did something wrong while row reducing the matrix equation associated with the above system of equations) to
$\begin{align*}
a+b+c+d = 0  \tag{i}\\
a+b - c -d= 0 \tag{ii}\\
a-b+c-d=0 \tag{iii}
\end{align*}$
Adding (i) and (ii), we get
$2(a+b)=0 \implies a+b=0$
Substituting $a+b=0$ in (i) we get $c+d=0$
Adding (ii) and (iii), we get
$2(a-d)=0 \implies a-d=0$
Substituting $a-d=0$ in (ii) we get $b-c=0$
Thus $a+b=0, c+d=0,a-d=0,b-c=0$ (You can get some more, but these are enough),
Taking $a=t$ gives $b=-a=-t, c=b=-t, d=a=t$
$\therefore u_4 = [a, b, c, d]=[t,-t,-t,t]$
