I forgot what was the answer given in class, but can we adjust the norm of (n x u) and add it to u to get an arbitrary angle?

Actually, you don't even need to adjust the norm: to rotate $u$ by an angle $\theta$ you can just write

$v = \cos(\theta) u + \sin(\theta)(n \times u).$

This vector will have the same norm as $u$, since

$|v|^2 = |\cos(\theta)u|^2 + 2\langle \cos(\theta) u, \sin(\theta)(n \times u) \rangle + |\sin(\theta)(n \times u)|^2 = (\cos(\theta)^2 + \sin(\theta)^2)||u|^2.$

The reason the cross-term disappears is that $u$ and $n \times u$ are orthogonal. The reason we can combine the $\cos(\theta)$ and $\sin(\theta)$ term is that $u$ and $n \times u$ have the same norm, even though they're not the same vector.

I forgot what was the answer given in class, but can we adjust the norm of (n x u) and add it to u to get an arbitrary angle?

Actually, you don't even need to adjust the norm: to rotate $u$ by an angle $\theta$ you can just write

$v = \cos(\theta) u + \sin(\theta)(n \times u).$

This vector will have the same norm as $u$, since

$|v|^2 = |\cos(\theta)u|^2 + 2\langle \cos(\theta) u, \sin(\theta)(n \times u) \rangle + |\sin(\theta)(n \times u)|^2 = (\cos(\theta)^2 + \sin(\theta)^2)||u|^2.$

The reason the cross-term disappears is that $u$ and $n \times u$ are orthogonal. The reason we can combine the $\cos(\theta)$ and $\sin(\theta)$ term is that $u$ and $n \times u$ have the same norm, even though they're not the same vector.