  We consider the following $q$-analog of the basic combinatorial search
problem: let $q$ be a prime power and $\GF(q)$ the finite field of $q$
elements. Let $V$ denote an $n$-dimensional vector space over $\GF(q)$ and let
$\mathbf{v}$ be an unknown 1-dimensional subspace of $V$. We will be interested
in determining the minimum number of queries that is needed to find
$\mathbf{v}$ provided all queries are subspaces of $V$ and the answer to a
query $U$ is YES if $\mathbf{v} \leqslant U$ and NO if $\mathbf{v}
\not\leqslant U$. This number will be denoted by $A(n,q)$ in the adaptive case
(when for each queries answers are obtained immediately and later queries might
depend on previous answers) and $M(n,q)$ in the non-adaptive case (when all
queries must be made in advance).
  In the case $n=3$ we prove $2q-1=A(3,q)<M(3,q)$ if $q$ is large enough. While
for general values of $n$ and $q$ we establish the bounds \[ n\log q \le A(n,q)
\le (1+o(1))nq \] and \[ (1-o(1))nq \le M(n,q) \le 2nq, \] provided $q$ tends
to infinity.
