An experimenter's logbook contains the following entry: "a charged...

The magnetic force on a moving charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$.

Since the magnetic force is always perpendicular to the magnetic field, the acceleration $\vec{a}$ of the particle must also be perpendicular to $\vec{B}$.

Therefore, the dot product of acceleration and magnetic field must be zero, $\vec{a} \cdot \vec{B} = 0$.

Given $\vec{B} = (7.0 \hat{i} - 3.0 \hat{j}) \times 10^{-3} \text{ T}$ and $\vec{a} = (x \hat{i} + 7.0 \hat{j}) \times 10^{-6} \text{ m s}^{-2}$.

Taking the dot product:

$(x \hat{i} + 7.0 \hat{j}) \cdot (7.0 \hat{i} - 3.0 \hat{j}) = 0$

$7.0x - 21.0 = 0$

$x = 3.0$

The missing number is $3.0$.