Find a vector orthogonal to both $(1,2,-1,3)$ and $(1,0,-2,3)$ So with my previous knowledge that the dot product of two orthogonal vectors should be equal to zero, I was wondering if the vector $(0,0,0,0)$ in $\mathbb{R}^4$ can be considered a vector orthogonal to the two vectors above, since its dot product with any other vector in $\mathbb{R}^4$ would be zero. Is my approach or intuition correct or logical?
 A: Yes, technically, the zero vector is orthogonal to both of your vectors. If this is a homework question, though, I'm certain that your professor intended for you to find a non-zero vector orthogonal to the two given vectors (which you can do fairly easily in this case by eyeballing the vectors, or more systematically using  e.g. Gram-Schmidt).
A: You can use Gram-Schmidt as suggested by @user7530. But you can also observe that any vector of the form $(2a,-a,3b,b)$ for $a,b\in\mathbb R$ is orthogonal to $(1,2,-1,3)$. And the dot-product of $(2a,-a,3b,b)$ with $(1,0,-2,3)$ is $2a-3b$. So choosing $a=3$ and $b=2$ gives $(6,-3,6,2)$.

Hope this helps.
A: Call the to-be-found-unknown-vector orthogonal to given vectors as $(x,y,z,w)$. Being orthogonal to a specific vector, say, $(a,b,c,d)$ means $ax+by+cz+dw=0$, (dot-product condition) and one more equation of the same kind for another orthogonality.
The problem boils down to solving a system of 2 equations in 4 variables; being homogeneous, zero vector is a solution, but as another answer points out, most likely you are expected to  find a non-zero solution.
A: Also note that you can solve this purely "3-dimensionally". Using the cross-product for vectors in $\mathbb{R}^3$, you see that $(-4,1,-2)$ is orthogonal to both $(1,2,-1)$ and $(1,0,-2)$. Now clearly $(-4,1,-2,0)$ is a solution to your problem.
A: The null vector is not a good choice since it is trivially orthogonal to every vector and no so much useful (it should be excluded by the problem set).
The problem to solve is to find a vector $v_3(a,b,c,d)$ such that $v_3\cdot v_1=0$ and $v_3 \cdot v_2=0$, that is equivalent to solve the system
$$Mv_3=\begin{bmatrix}
1 & 2 & -1&3\\
1 & 0 & -2&3
\end{bmatrix}\cdot\begin{bmatrix}
a\\b\\c\\
d
\end{bmatrix}=\begin{bmatrix}
0\\0\\0\\0
\end{bmatrix}$$
which has $\infty^2$ solutions.
You can also solve by inspection choosing a first vector orthogonal to $v_1$ keeping some free parameter and then making it orthogonal also to $v_2$, for example
$$v_3=(-2x,x,3y,y) \implies v_3\cdot v_1=0$$
then we need
$$v_3\cdot v_2=0\implies -2x-6y+3y=0\implies 2x=-3y \implies x=3 \quad y=-2 $$
thus
$$v_3=(-6,3,-6,-2)$$
