Answer
\[
\vec{v} \times \vec{u}=\hat{\imath}+2 \hat{\jmath}-4 \hat{k}
\]
We have proven that $\vec{v}$ , $\vec{u}$ are cross-referenced with their cross product $\vec{v} \times \vec{u}$
Work Step by Step
For two 3-Dimensional vectors:
\[
\vec{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\quad \text { and } \quad \vec{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle
\]
The cross product is given by the determinant
\[
\vec{v} \times \vec{u}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3}
\end{array}\right|
\]
And then, for $\vec{v}=2 \hat{\imath}-\hat{\jmath}=\langle 2,-1,0\rangle$ and $\vec{u}=4 \hat{\imath}+\hat{k}=\langle 4,0,1\rangle$, we get:
\[
\begin{aligned}
\vec{u} \times \vec{v} &=\left|\begin{array}{ccc}
1 & \hat{\jmath} & \hat{k} \\
4 & 0 & 1 \\
2 & -1 & 0
\end{array}\right|=\hat{\imath}(0+1)-\hat{\jmath}(0-2)+\hat{k}(-4+0) \\
&=\hat{\imath}+2 \hat{\jmath}-4 \hat{k}=\langle 1,2,-4\rangle
\end{aligned}
\]
Notice that:
\[
(\vec{v} \times \vec{u}) \cdot\vec{u}=\langle 1,2,-4\rangle\langle 4,0,1\rangle=-4+4+0=0
\]
\[
(\vec{v} \times \vec{u})\cdot\vec{v} =\langle 1,2,-4\rangle\langle 2,-1,0\rangle=-2+2+0=0
\]
So, $\vec{v}$ and $\vec{u}$ are orthogonal to the cross product $\vec{v} \times \vec{u}$
