The Electric field of a plane electromagnetic wave is given by$\overrightarrow{E}=E_0\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\cos \left(Kz+\omega t\right)$. At t=0, a positively charged particle is at the point  $\left(x,y,z\right)=\left(0,0,\frac{\pi }{K}\right)$.If its instantaneous velocity at (t=0) is  $v_0\widehat{k}$ , the force acting on it due to the wave is:

  $\begin{array}{l}At t=0, x =K \text{}\\ Electric field =\overrightarrow{E}=E_0\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt{2}}\cos \left(\pi \right)=-\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt{2}}{E}_0\\ \end{array}$

$\therefore$  $Force due to electric field is in direction -\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt{2}}$

$\begin{array}{l}Direction ofelectromagnetic wave is in \left(-z\right) direction\text{}\\ \therefore \widehat{C}=-\widehat{k} is wave direction.\\ Let \widehat{B} is the unit vector along the direction of magnetic field.\text{} \\ \widehat{B}=\widehat{C}\times \widehat{E} = -\widehat{k}\times -\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt{2}}=\frac{\left(-\widehat{i}+\widehat{j}\right)}{\sqrt{2}}\end{array}$
Now, Force due to magnetic field is in direction $q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$  and  $\overrightarrow{v}\left|\right|\widehat{k}$

$Force due to magnetic field is \hat{F}=\hat{k}\times \left(\frac{-\hat{i}+\hat{j}}{\sqrt{2}}\right) =\left(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\right)=-\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt{2}}$
$We can see that direction of both electric and magnetic force is same.$
$\therefore$  Net force is antiparallel to  $\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt{2}}$