Motion in a Plane - Result Question 3
3. 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 :
[2015 RS]
(a) $t=\frac{\pi}{2 \omega}$
(b) $t=\frac{\pi}{\omega}$
(c) $t=0$
(d) $t=\frac{\pi}{4 \omega}$
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Answer:
Correct Answer: 3. (b)
Solution:
- (b) Two vectors are
$\overrightarrow{{}A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$
$\overrightarrow{{}B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}$
For two vectors $\vec{A}$ and $\vec{B}$ to be orthogonal $\vec{A} \cdot \vec{B}=0$
$\overrightarrow{{}A} \cdot \overrightarrow{{}B}=0=\cos \omega t \cdot \cos \frac{\omega t}{2}+\sin \omega t \cdot \sin \frac{\omega t}{2}$
$ =\cos (\omega t-\frac{\omega t}{2})=\cos (\frac{\omega t}{2}) $
So, $\frac{\omega t}{2}=\frac{\pi}{2} \therefore t=\frac{\pi}{\omega}$
