Find an orthononormal basis of $U^{\bot}$ given $U$ Given $U =$ span$\lbrace u_1 = (6,2,-2,-2),u_2 = (-1,1,-1,-1)\rbrace$ find an orthonormal basis for $U$ and $U^\bot$
What I've done so far: $(<u_1,u_2>) = 0 \implies u_1\bot u_2 \implies$ orthonormal basis given by $e_1 = \frac{u_1}{||u_1||},e_2 = \frac{u_2}{||u_2||}$. 
However I dont know how to find the orthonormal basis for $U^\bot$ from here, I tried setting $v_1 = (a,b,c,d)$ and solving $(<v,u1>) = 0 \wedge(<v,u2>) = 0$ but only got $(a,b,c,d) = \vec{0}$ or $b = c + d$.
Thanks in advance :)

As you've noted, $e_1,e_2$ is an orthonormal basis for $U$. Now, that basis can be expanded with some $e_3,e_4$ to an orthonormal basis of $\Bbb R^4$ (I am assuming that we are working in $\Bbb  R^4$). Then $e_3$ and $e_4$ must be an orthonormal basis for $U^\perp$. Let's show this. First, $e_3$ and $e_4$ are linearly independent and orthonormal as they are part of an orthonormal basis for $\Bbb R^4$. Now, it is easy to see that $e_1\cdot(ae_3+be_4)=e_2 \cdot (ce_3+de_4)=0$ as $e_1,e_2,3_3,e_4$ are all orthogonal, so $\operatorname{span}\{e_3,e_4\}\subset U^\perp$. On the other hand, suppose $v\in U^\perp$. Then $v=a_1e_1+a_2e_2+a_3e_3+a_4e_4$ as they form a basis for $\Bbb R^4$. However, that means that $\langle e_1,v\rangle=a_1=0$ and $\langle e_2,v\rangle = a_2 = 0$, so $v=a_3e_3+a_4e_4\in\operatorname{span}\{e_3,e_4\}$. Therefore, $e_3$ and $e_4$ are a basis for $U^\perp$.
Now the question is simply how to extend $e_1$ and $e_2$ to an orthonormal basis. To do this, you can use the Gramm-Schmidt procedure after extending $e_1$ and $e_2$ to a basis for $\Bbb R^4$.
