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If vectors $\vec{A} = \cos \omega t\hat{i} + \sin \omega t\hat{j}$ and $\vec{B} = \cos (\omega t/2)\hat{i} + \sin (\omega t/2)\hat{j}$ are functions of time, then the value of t at which they are orthogonal to each other is:
Explanation
Two vectors are $\vec{A} = \cos \omega t\hat{i} + \sin \omega \hat{j}$; $\vec{B} = \cos(\omega t/2)\hat{i} + \sin(\omega t/2)\hat{j}$ For two vectors $\vec{A}$ and $\vec{B}$ to be orthogonal $\vec{A} \cdot \vec{B} = 0$ $\vec{A} \cdot \vec{B} = 0 = \cos \omega t \cdot \cos(\omega t/2) + \sin \omega t \cdot \sin(\omega t/2)$ = $\cos(\omega t - \omega t/2) = \cos(\omega t/2)$ So, $\omega t/2 = \pi/2 \therefore t = \pi/\omega$