Finding an orthogonal vector to two vectors in $\mathbb{R}^4$ "Let $u_1$, $u_2$ be to vectors in $\mathbb{R}^4$ $$u_1=(1,0,1,1) \text{ and } u_2=(1,1,0,3)$$
Provide a real vector which is orthogonal to both $u_1$ and $u_2$
So, I kind of guessed a vector $u_3=(1,-1,-1,0)$ which must be orthogonal to both since $$u_1 \cdot u_3 = 0 \text{ and } u_2 \cdot u_3=0$$
My question is, how should it be done if it can't immediately be guessed? In $\mathbb{R}^3$ one could just take the cross product of the two vectors, but that's not defined for any other vector spaces
 A: You can use Gauss-Jordan method to solve the linear system:
$$\langle (x,y,z,t),u_1\rangle=0$$
$$\langle (x,y,z,t),u_2\rangle=0$$
that is:
$$ x+z+t=0 $$
$$ x+y+3t=0$$
so the matrix of your system of linear equations is:
$$
A=\left[
\begin{array}{cccc}
1&0&1&1\\
1&1&0&3
\end{array}
\right]
$$
the rref of $A$ is 
$$
\mathrm{rref}(A)
=
\left[
\begin{array}{cccc}
1&0&1&1\\
0&1&-1&2
\end{array}
\right]
$$
so, you have
$$x=-z-t$$
$$y=z-2t$$
with $z,t$ being any real values.
then all vectors $\vec{v}=(x,y,z,t)$ that are ortogonal to both $u_1$ and $u_2$ are of the form
$$\vec{v}=(-z-t,z-2t,z,t).$$

See the link bellow to clarify the general cross product:
Cross product in $\mathbb R^n$

A: Take a non-null vector $u_3$. If it is orthogonal to the other two, you're done. Otherwise, compute the $4$-dimensional cross-product of $u_1$, $u_2$, and $u_3$ described here.
A: Since you have only two vectors, you can work in $\mathbb R^3$. Let $u_3 = (a, b, c, 0)$, so that $u_1 \cdot u_3$ and $u_2 \cdot u_3$ only depend on the first three components of $u_1$ and $u_2$. So, call $v_i$ the vector of the first three components of $u_i$, and compute $v_3 = v_1 \times v_2 = (a, b, c)$ (this wouldn't work if either $v_1$ or $v_2$ was zero).
Specifically,
$$u_3 = \begin{pmatrix}
\begin{vmatrix}0 & 1 \\ 1 & 0\end{vmatrix},
-\begin{vmatrix}1 & 1 \\ 1 & 0\end{vmatrix},
\begin{vmatrix}1 & 0 \\ 1 & 1\end{vmatrix},
0
\end{pmatrix} = (-1, 1, 1, 0)$$
Which only differs in sign from your solution.
