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If vectors $\vec A = \cos\omega t\,\hat i + \sin\omega t\,\hat j$ and $\vec B = \cos\frac{\omega t}{2}\,\hat i + \sin\frac{\omega t}{2}\,\hat j$ are functions of time, then the value of t at which they are orthogonal to each other is:
A
t = 0
B
$t = \frac{\pi}{4\omega}$
C
$t = \frac{\pi}{2\omega}$
D
$t = \frac{\pi}{\omega}$
Detailed Solution
Orthogonal: $\vec A\cdot\vec B = 0$
$\cos\omega t\cos\frac{\omega t}{2} + \sin\omega t\sin\frac{\omega t}{2} = 0 \Rightarrow \cos\left(\omega t - \frac{\omega t}{2}\right) = \cos\frac{\omega t}{2} = 0$
$\frac{\omega t}{2} = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{\omega}$
$\cos\omega t\cos\frac{\omega t}{2} + \sin\omega t\sin\frac{\omega t}{2} = 0 \Rightarrow \cos\left(\omega t - \frac{\omega t}{2}\right) = \cos\frac{\omega t}{2} = 0$
$\frac{\omega t}{2} = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{\omega}$