If vectors $$A = \cos \omega t\,\hat i + \sin \omega t\,\hat j$$     and $$B = \cos \frac{{\omega t}}{2}\,\hat i + \sin \frac{{\omega t}}{2}\,\hat j$$     are functions of time, then the value of $$t$$ at which they are orthogonal to each other, is

Solution :
For perpendicular vector, we have $$A \cdot B = 0$$
$$\eqalign{ & \left[ {\cos \omega t\,\hat i + \sin \omega t\,\hat j} \right]\left[ {\cos \frac{{\omega t}}{2}\hat i + \frac{{\sin \omega t}}{2}\hat j} \right] = 0 \cr & \Rightarrow \cos \omega t\cos \frac{{\omega t}}{2} + \sin \omega t\,\sin \frac{{\omega t}}{2} = 0\,\left[ {\because \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B} \right] \cr & \Rightarrow \cos \left( {\omega t - \frac{{\omega t}}{2}} \right) = 0 \Rightarrow \cos \frac{{\omega t}}{2} = 0 \cr & \Rightarrow \frac{\omega }{2} = \frac{\pi }{2} \Rightarrow t = \frac{\pi }{\omega } \cr} $$
Thus, time taken by vectors which are orthogonal to each other is $$\frac{\pi }{\omega }.$$