Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are $\overrightarrow{v}=3\hat{i}+4\hat{j}$ and $\overrightarrow{a}=2\hat{i}+x\hat{j}$ (Select the correct alternative(s).

The correct options are
A $x=-1.5$
C Magnetic field is along z - direction.
D Kinetic energy of the particle is constant
The magnetic force $F_m$ is given by
$\overrightarrow{F_m}=q\overrightarrow{V}\times \overrightarrow{B}$
$⟹m\overrightarrow{a}=q\overrightarrow{V}\times \overrightarrow{B}$---(i)

Using the above equation we can say $\overrightarrow{a}$ is perpendicular to both $\overrightarrow{V}$ and $\overrightarrow{B}$.

So, $\overrightarrow{a}.\overrightarrow{V}=0$
$⟹(2\hat{i}+x\hat{j}).(3\hat{i}+4\hat{j})$ = 0
$⟹6+4x=0$
$⟹x=\frac{-3}{2}=-1.5$
Both $\overrightarrow{V}$ and $\overrightarrow{a}$ are in $x-y$ plane so $\overrightarrow{B}$ must be in $z$ direction to satisfy equation (i).
As mass and magnitude of velocity both are constant, it means kinetic energy will also be constant.