Vectors in three dimensions and two unknown components
Solution 1:
The usual inner product on $\mathbb{R}^{3}$ is given by: $$\langle v, w \rangle := v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3}$$ where $v = (v_{1},v_{2},v_{3})$ and $w= (w_{1},w_{2},w_{3})$. In your case: $$\langle v, w \rangle = \frac{1}{4}+\frac{b}{3}-a.$$ As you mentioned, in order to $v$ and $w$ to be orthogonal we must have: $$\langle v, w \rangle = 0$$ so that any $a$ and $b$ satisfying: $$\frac{1}{4}+\frac{b}{3}-a = 0$$ fulfills the condition. We have infinitely many answers and you can pick any of such solutions as the solution of your problem.