I already have two vectors expressed in basis $\{|u_1 \rangle, |u_2\rangle, |u_3\rangle\}$:
$$|\phi_1\rangle = \frac{1}{\sqrt{3}}(|u_1\rangle + i |u_2\rangle - |u_3\rangle) $$
$$|\phi_2\rangle = \frac{1}{\sqrt{2}}(|u_1\rangle - i |u_2\rangle) $$
and I have to find a third one to form an orthonormal basis. How do I do that ?
I tried using the relations :
$$\langle\phi_1\mid\phi_3\rangle = 0$$
$$\langle\phi_2\mid\phi_3\rangle= 0$$
$$\langle\phi_3\mid\phi_3\rangle = 1$$
Which leads to :
$$x_3 = 2x_1$$
$$x_2 = ix_1$$
$$x_1^{*}x_1 + x_2^{*}x_2 + x_3^{*}x_3 = 1$$
where $x_1, x_2, x_3$ are the components of the third vector. But I can't seem to go further than that...
First remember that $\{|u_1\rangle, |u_2\rangle, |u_3\rangle\} \not\subseteq \operatorname{span}(|\phi_1\rangle, |\phi_2\rangle)$ -- in fact the span might not include any of them. So just start checking whether each of your basis vectors is in the span and stop when you find one that isn't.
Clearly $|u_1\rangle \not\in \operatorname{span}(|\phi_1\rangle, |\phi_2\rangle)$. So, because this Hilbert space is $3$-dimensional, $\{|u_1\rangle, |\phi_1\rangle, |\phi_2\rangle\}$ is a basis for the space.
Note that this is not an orthonormal basis. But you can easily make it into one via the Gram-Schmidt process.