= Solution
Take the scalar product of the <Lorentz force> equation with the constant <magnetic field> $\mathbf B$. Since $\mathbf E\cdot\mathbf B=0$ and $(\dot{\mathbf x}\times\mathbf B)\cdot\mathbf B=0$,
$$
m\frac{d^2}{dt^2}(\mathbf x\cdot\mathbf B)=0.
$$
The initial velocity also obeys $\mathbf v\cdot\mathbf B=0$, so $\mathbf x\cdot\mathbf B$ remains equal to $\mathbf x_0\cdot\mathbf B$. The trajectory therefore lies in the plane
$$
\boxed{(\mathbf x-\mathbf x_0)\cdot\mathbf B=0},
$$
the plane through $\mathbf x_0$ perpendicular to $\mathbf B$.
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