d
\(\begin{gathered}
  \,\,\,\,\,Two\,vectors\,\bar A\,and\,\bar B\,are\,orthogonal \hfill \\
  to\,each\,other,\,if\,their\,scalar\,product \hfill \\
  is\,zero\,i.e,.\,\bar A\,.\,\bar B = 0. \hfill \\
  Here,\,\bar A = \cos \omega t\,\hat i + \sin \omega t\,\hat j \hfill \\
  \,\,\,\,\,\,\,\,\,and\,\bar B = \cos \frac{{\omega t}}{2}\hat i + \sin \frac{{\omega t}}{2}\hat j \hfill \\
  \therefore \,\bar A\,.\,\bar B = \left( {\cos \omega t\hat i + \sin \,\omega t\,\hat j} \right) \cdot \left( {\cos \frac{{\omega t}}{2}\hat i + \sin \frac{{\omega t}}{2}\hat j} \right) \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\cos \,\omega t\,\cos \,\frac{{\omega t}}{2} + \sin \,\omega t\,\sin \,\frac{{\omega t}}{2} \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \,\hat i \cdot \hat i = \hat j \cdot \hat j = 1\,and\,\hat i \cdot \hat j = \hat j \cdot \hat i = 0} \right) \hfill \\ 
\end{gathered} \)

\(\begin{gathered}
  \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \cos \left( {\omega t - \frac{{\omega t}}{2}} \right) \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B} \right) \hfill \\
  But\,\bar A\,.\,\bar B = 0\left( {as\bar A\,\,and\,\bar B\,are\,orthogonal\,to\,each\,other} \right) \hfill \\
  \therefore \,\,\,\cos \left( {\omega t - \frac{{\omega t}}{2}} \right) = 0 \hfill \\
  \,\,\,\,\,\,\cos \left( {\omega t - \frac{{\omega t}}{2}} \right) = \cos \frac{\pi }{2}\,or\,\omega t - \frac{{\omega t}}{2} = \frac{\pi }{2} \hfill \\
  \,\,\,\,\,\,\frac{{\omega t}}{2} = \frac{\pi }{2}or\,t = \frac{\pi }{\omega } \hfill \\ 
\end{gathered} \)