A positively charged particle with velocity $v\hat k$ moves through an unknown (but uniform) magnetic field. The force on the particle is observed to be $\vec F = F_0 (3 \hat{i} + 4\hat j)$. What is the $x$-component of the magnetic field?
The equation we use to solve these problems: $\vec F = q\vec v \times \vec B$
So I know $\vec F$, so I plug that in: $F_0 (3 \hat{i} + 4\hat j) = v\hat k \times \vec B$.
How do I proceed to get the $x$-component?
Note: The answer to this problem is $\frac{4F_0}{qv}$
You have that $$ F_0\,\left(3\hat{i}+4\hat{j}\right) = q\, v\,\hat{k}\times\vec{B}, $$ which means that $$ F_0\,\left(3\hat{i}+4\hat{j}\right) = q\, v\,\hat{k}\times\left(B_x\hat{i}+B_y\hat{j}+B_z\hat{k}\right), $$ therefore, recalling that $\hat{k}\times\hat{i}=\hat{j}$ and that the corresponding components of the vectors shall be equal, you get: $$ 4\,F_0\hat{j}=q\,v\,B_x\,\left(\hat{k}\times\hat{i}\right), $$ i.e. $$ 4\,F_0\hat{j}=q\,v\,B_x\,\hat{j}, $$ hence $$ 4\,F_0=q\,v\,B_x\, $$ which results precisely in $$ B_x={4\,F_0\over q\,v}. $$