Just wanna see if this is on the right track? Or if I have been pointed in the wrong direction?
Started with vector cross product:
$$\textbf{V}\times\textbf{U}=(-m-3)\textbf{i}+8\textbf{j}+(2-2m)\textbf{k}.$$
Found the magnitude of $\textbf{V}$ and $\textbf{U}:$
$$||\textbf{V}|| = \sqrt{13+m^2}\ \text{and}\ ||\textbf{U}||=\sqrt6.$$
Was told to do dot product using the magnitudes found. Perpindicular is $90°$ and Parallel is $180°$.
Applying $\textbf{V}\times\textbf{U}=||\textbf{V}||\cdot||\textbf{U}||\sin\theta$ with $\theta=180°$ I got $m = -8$. But other then that I'm not sure if I've gone on the right path or where to go from here.
Quite confused with the wording of the question, any help greatly appreciated.
The dot product would help you more, as the cross product is used to find a vector at a right angle to both $\vec{v}$ and $\vec{u}$.
Because the first vector has a variable $m$, it's best to start with the second and find a parallel vector. Remember there are two parallel unit vectors to $\vec{u}$; one in its direction and the other in the negative.
$ \left| \vec{u} \right| = \sqrt{(2)^2+(1)^2+(-1)^2} = \sqrt{6}$, then $ \hat{u} = \frac{1}{\sqrt6}\vec{u} = \frac{2\sqrt{6}}{6}\tilde{i}+\frac{\sqrt{6}}{6}\tilde{j}-\frac{\sqrt{6}}{6}\tilde{k}$. Multiplying this by $-1$ gives you the parallel vector in the opposite direction if needed, but the question only asks for one example so in this instance it's not essential.
To find a vector perpendicular to $\vec{v}$, $ \vec{v} \cdot \hat{u} = 0$, then
$\vec{v}\perp = (2)(\frac{2\sqrt{6}}{6}) + (m)(\frac{\sqrt{6}}{6})+(-3)(-\frac{\sqrt{6}}{6}) = 0$. Solving for $m$ it can be concluded that $m=-7$, so the resulting vector is $\vec{v}\perp = 2\tilde{i} -7\tilde{j} +3\tilde{k}$
Then construct the unit vector using $ \left| \vec{v}\perp \right| = \sqrt{(2)^2+(-7)^2+(3)^2} = \sqrt{62}$
$\hat{v}\perp = \frac{1}{\sqrt{62}}(2\tilde{i} -7\tilde{j} +3\tilde{k})$. Hope that makes sense