Problem statement:
Find all vectors in $\mathbb R^4$ that are orthogonal to the two vectors
$u_1=(1,2,1,3)$ and $u_2=(2,5,1,4)$.
Progress:
a) Denote a vector $u_3=(v_1,v_2,v_3,v_4)$ My desire is to determine $u_3$ so that $\left \langle u_1,u_3 \right \rangle=\left \langle u_2,u_3 \right \rangle=0$
$\left \langle u_1,u_3 \right \rangle=(1,2,1,3)*(v_1,v_2,v_3,v_4)=v_1+2v_2+v_3+3v_4=0$
$\left \langle u_2,u_3 \right \rangle=(2,5,1,4)*(v_1,v_2,v_3,v_4)=2v_1+5v_2+v_3+4v_4=0$
Thus I end up with a system of equations I could not solve.
b) I know that the cross product of two vectors $a$ and $b$ results in a vector orthogonal to $a$ and $b$ that cannot be applied in $\mathbb R^4$.
Best Answer
For a), you have the correct system of equations. Do you know the Gauss-Jordan elimination algorithm? When you apply it to your system of equations, you will find that there are two free variables, so you will have a two-dimensional subspace.
For b), you are correct that the cross-product is only defined for vectors in $\mathbb{R}^3$.
Intuitively, to be orthogonal to the given vectors $u_1$ and $u_2$ is equivalent to being orthogonal to the plane spanned by $u_1$ and $u_2$. In $\mathbb{R}^4$, the set of all vectors orthogonal to a plane has dimension $4 - 2 = 2$, so it is also a plane.
SPOILER: To check your work, the canonical basis obtained by solving your system of equations is $w_1 = (-3,1,1,0)$ and $w_2 = (-7,2,0,1)$. I'd be happy to give more details if you're stuck, but it sounds like you are close to solving it on your own.