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(D) Two vectors $\overrightarrow {A}$ and $\overrightarrow {B}$ are orthogonal to each other if their scalar product is zero,i.e.,$\overrightarrow {A}

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$t=\frac{\pi}{\omega}$

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(D) Two vectors $\overrightarrow {A}$ and $\overrightarrow {B}$ are orthogonal to each other if their scalar product is zero,i.e.,$\overrightarrow {A} \cdot \overrightarrow {B} = 0$.Given $\overrightarrow {A} = \cos \omega t \hat i + \sin \omega t \hat j$ and $\overrightarrow {B} = \cos \frac{\omega t}{2} \hat i + \sin \frac{\omega t}{2} \hat j$.Calculating the dot product:$\overrightarrow {A} \cdot \overrightarrow {B} = (\cos \omega t \hat i + \sin \omega t \hat j) \cdot (\cos \frac{\omega t}{2} \hat i + \sin \frac{\omega t}{2} \hat j)$$= \cos \omega t \cos \frac{\omega t}{2} + \sin \omega t \sin \frac{\omega t}{2}$Using the trigonometric identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$\overrightarrow {A} \cdot \overrightarrow {B} = \cos(\omega t - \frac{\omega t}{2}) = \cos(\frac{\omega t}{2})$Since the vectors are orthogonal,$\overrightarrow {A} \cdot \overrightarrow {B} = 0$,so:$\cos(\frac{\omega t}{2}) = 0$We know $\cos(\frac{\pi}{2}) = 0$,therefore:$\frac{\omega t}{2} = \frac{\pi}{2}$$t = \frac{\pi}{\omega}$.

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

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