if $\vec{x},\vec{y} \in \Bbb{R^4}$ with $\vec{x},\vec{y}$ being linearly independent prove that there exists a $\vec{a}$ that is orthogonal to both?

Question

let $\vec{x},\vec{y} \in \Bbb{R^4}$ with $\vec{x},\vec{y}$ being linearly independent how would you prove that there exists a non-zero vector $\vec{a}$ that is orthogonal to both?

I know that since its linearly independent the only solution to $c_1\vec{x}+c_2\vec{y}=0$ is when $c_1=c_2=0$ and that $\vec{a}\cdot\vec{x}=0$, $\vec{a}\cdot\vec{y}=0$ but im stuck on how to use all this to reach the conclusio

Answer 1

Since $\vec x,\vec y$ are linearly independent, their span forms a $2$-dimensional plane in $\Bbb R^4$. Consider a normal vector to this plane.

Answer 2

The null space of a matrix is the orthogonal complement of its row space, so you can take any vector in the null space of $\small\begin{bmatrix}\vec x & \vec y \end{bmatrix}$.