A) \[t=\frac{\pi }{4\omega }\]
B) \[t=\frac{\pi }{2\omega }\]
C) \[t=\frac{\pi }{\omega }\]
D) \[t=0\]
Correct Answer: C
Solution :
| For perpendicular vector, we have \[\mathbf{A}.\mathbf{B}=0\] |
| \[[cos\omega t\hat{i}+sin\omega t\hat{j}].\left[ \cos \frac{\omega t}{2}\hat{i}+\frac{\sin \omega t}{2}\hat{j} \right]=0\] |
| \[\Rightarrow \] \[\cos \omega t\cos \frac{\omega t}{2}+\sin \omega t.\sin \frac{\omega t}{2}=0\] |
| \[[\because \,\cos (A-B)=\cos A\cos B+\sin A\sin B]\] |
| \[\Rightarrow \] \[\cos \left( \omega t-\frac{\omega t}{2} \right)=0\] |
| \[\cos \frac{\omega t}{2}=0\Rightarrow \frac{\omega t}{2}=\frac{\pi }{2}\Rightarrow t-\frac{\pi }{\omega }\] |
| Thus, time taken by vectors which are orthogonal, to each other is \[\frac{\pi }{\omega }\]. |
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