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

$\overrightarrow{\mathrm{A}}=\mathrm{cos\omega t}\widehat{\mathrm{i}}+\mathrm{sin\omega t}\widehat{\mathrm{j}}$

$\overrightarrow{\mathrm{B}}=\cos \left(\frac{\mathrm{\omega t}}{2}\right)\widehat{\mathrm{i}}+\sin \left(\frac{\mathrm{\omega t}}{2}\right)\widehat{\mathrm{j}}$

For $\overrightarrow{\mathrm{A}}\mathrm{and}\overrightarrow{\mathrm{B}}\mathrm{orthogonal}\overrightarrow{\mathrm{A}.}\overrightarrow{\mathrm{B}}=0$

$\left(\mathrm{cos\omega t}\widehat{\mathrm{i}}+\mathrm{sin\omega t}\widehat{\mathrm{j}}\right).\left(\cos \frac{\mathrm{\omega t}}{2}\widehat{\mathrm{i}}+\sin \frac{\mathrm{\omega t}}{2}\widehat{\mathrm{j}}\right)=0$

$\mathrm{cos\omega t}.\cos \frac{\mathrm{\omega t}}{2}+\mathrm{sin\omega t}.\sin \frac{\mathrm{\omega t}}{2}=0$

$\cos (\mathrm{\omega t}-\frac{\mathrm{\omega t}}{2}) =0$

$\Rightarrow \cos \frac{\mathrm{\omega t}}{2}=0$

$\frac{\mathrm{\omega t}}{2}=\frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{\omega t}=\mathrm{\pi }\Rightarrow \mathrm{t}=\frac{\mathrm{\pi }}{\mathrm{\omega }}$