Answer
\[
\pm\left\langle\frac{2}{\sqrt{30}}, \frac{5}{\sqrt{30}},-\frac{1}{\sqrt{30}}\right\rangle
\]
Work Step by Step
\[
\begin{array}{l}
\vec{u}=-7 \hat{\imath}+3 \hat{\jmath}+\hat{k}=\langle-7,3,1\rangle \\
\vec{v}=2 \hat{\imath}+4 \hat{k}=\langle 2,0,4\rangle
\end{array}
\]
Since $\vec{w}=\vec{v} \times \vec{u}$ is orthogonal to both $(\vec{v} \text { , } \vec{u})$, we get that $-\frac{\vec{w}}{|\vec{v}|}$ and $\frac{\vec{w}}{|\vec{w}|}$ are unit vectors perpendicular to $\vec{v}$ and $\vec{u}$. We evaluate $\vec{v} \times \vec{u}=\vec{w}$
\[
\begin{aligned}
\vec{w}=\vec{u} \times \vec{v}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-7 & 3 & 1 \\
2 & 0 & 4
\end{array}\right| &=\hat{i}(12-0)+\hat{j}(2+28)+\hat{k}(0-6) \\
&=12 \hat{\imath}+30 \hat{j}-6 \hat{k} \\
&
\end{aligned}
\]
So,
\[
\begin{aligned}
\frac{\vec{w}}{\|\vec{w}\|} &=\langle 12,30,-6\rangle \frac{1}{6 \sqrt{30}} \\
&=\left\langle\frac{2}{\sqrt{30}}, \frac{5}{\sqrt{30}},-\frac{1}{\sqrt{30}}\right\rangle
\end{aligned}
\]
\[
\pm\left\langle\frac{2}{\sqrt{30}}, \frac{5}{\sqrt{30}},-\frac{1}{\sqrt{30}}\right\rangle
\]
They are unit vectors perpendicular to $\bar{v}$, $\vec{u}$.
