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

1. \( t = \frac{\pi}{2\omega} \)
2. \( t = \frac{\pi}{\omega} \)
3. \( t = 0 \)
4. \( t = \frac{\pi}{4\omega} \)

Correct answer: \( t = \frac{\pi}{\omega} \)

Solution

Two vectors are orthogonal if their dot product is zero. The dot product of \( \mathbf{A} \) and \( \mathbf{B} \) is \( \cos \omega t \cos \frac{\omega t}{2} + \sin \omega t \sin \frac{\omega t}{2} = \cos \left( \omega t - \frac{\omega t}{2} \right) = \cos \frac{\omega t}{2} \). For orthogonality, \( \cos \frac{\omega t}{2} = 0 \), which gives \( \frac{\omega t}{2} = \frac{\pi}{2} \), or \( t = \frac{\pi}{\omega} \).

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