On the method described by Purcell for finding the magnetic field by measuring the force on a test particle The following text is a method for finding the magnetic field as described in Purcell's Electricity and Magnetism (page $151$, the top part).

Measure the force on the particle when its velocity is $\bf{v}$; repeat with $\bf{v}$ in some other direction. Now find a $\bf{B}$ that will make $\textbf{f}=q\textbf{v}\times\textbf{B}$ fit all these results.

Why do we need two velocity vectors? 
For a given velocity $\textbf{v}_{1}=[v_{1} v_{2} v_{3}]^{T}$, there will be a corresponding force $\textbf{f}=[f_{1} f_{2} f_{3}]^{T}$. Therefore, the equation $\textbf{f}=q\textbf{v}\times\textbf{B}$ will be equivalent to 
$$f_{1}=q(v_{2}B_{3}-v_{3}B_{2})$$ 
$$f_{2}=q(v_{1}B_{3}-v_{3}B_{1})$$ 
$$f_{3}=q(v_{1}B_{2}-v_{2}B_{1})$$ 
Three equations, three unknowns. Am I missing something?
 A: The system of equations
$$f_{1}=q(v_{2}B_{3}-v_{3}B_{2})$$
$$f_{2}=q(v_{1}B_{3}-v_{3}B_{1})$$
$$f_{3}=q(v_{1}B_{2}-v_{2}B_{1})$$ 
Can be rewritten as $$\mathbf{f}=q\mathbf{M} \mathbf{B}$$ where
$$\mathbf{M}=\left(
\begin{array}{ccc}
 0 & -v_3 & v_2 \\
 v_3 & 0 & -v_1 \\
 -v_2 & v_1 & 0 \\
\end{array}
\right)$$
Note that $\det(\mathbf{M})=0$ so the system is not linearly independent. So even though you do have three equations in three unknowns there is not a unique solution. You need at least two non-zero $\mathbf{v}$ to solve it.
A: The linear system you've written,
$$
\vec f=
\begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix}
=q
\begin{pmatrix}
  0  & -v_3 &  v_2 \\
 v_3 &   0  & -v_1 \\
-v_2 &  v_1 &   0
\end{pmatrix}
\begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix}
=q(\vec v\times) \vec B
$$
is indeterminate, and it does not have a unique solution. You can easily verify this by noticing that the determinant is zero, since the matrix $M=(\vec v\times)$ is antisymmetric, but the transpose can't change the determinant, so
$$
\det(M) = \det(M^T) = \det(-M) = (-1)^3\det(M) = -\det(M),
$$
which requires $\det(M)=0$. 
Alternatively, you can just notice that it is unable to distinguish a magnetic field parallel to the velocity from a vanishing field (since both give zero force). This is seen most simply by choosing your coordinate axes so that $\vec v$ lies along the $x$ axis, so that the system of equations reads
$$
\begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix}
=
qv_1
\begin{pmatrix}
  0  & 0 &  0 \\
 0 &   0  & -1 \\
0 &  1 &   0
\end{pmatrix}
\begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix},
$$
which is clearly sufficient to figure out $B_2$ and $B_3$ from $f_2$ and $f_3$, but cannot say anything about $B_1$.
