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Step-by-Step Solution
Step 1: Understanding Orthogonality of Vectors
Two vectors are orthogonal if and only if their dot product equals zero. Mathematically, if
$ \overrightarrow{A} \cdot \overrightarrow{B} = 0 $, then $ \overrightarrow{A} $ and $ \overrightarrow{B} $ are perpendicular to each other.
Step 2: Write Down the Given Vectors
We have:
$ \overrightarrow{A} = \cos(\omega t)\,\hat{i} + \sin(\omega t)\,\hat{j} $
and
$ \overrightarrow{B} = \cos\Bigl(\frac{\omega t}{2}\Bigr)\,\hat{i} + \sin\Bigl(\frac{\omega t}{2}\Bigr)\,\hat{j}. $
Step 3: Compute the Dot Product
The dot product $ \overrightarrow{A} \cdot \overrightarrow{B} $ is:
\[
\bigl(\cos(\omega t)\,\hat{i} + \sin(\omega t)\,\hat{j}\bigr)
\,\cdot\,
\bigl(\cos\bigl(\tfrac{\omega t}{2}\bigr)\,\hat{i} + \sin\bigl(\tfrac{\omega t}{2}\bigr)\,\hat{j}\bigr).
\]
Since $ \hat{i}\cdot\hat{i} = 1 $, $ \hat{j}\cdot\hat{j} = 1 $, and $ \hat{i}\cdot \hat{j} = 0 $, this becomes:
\[
\cos(\omega t)\,\cos\Bigl(\tfrac{\omega t}{2}\Bigr) + \sin(\omega t)\,\sin\Bigl(\tfrac{\omega t}{2}\Bigr).
\]
Step 4: Simplify Using a Trigonometric Identity
Recall the identity:
\[
\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B).
\]
Here, if we let $ A = \omega t $ and $ B = \tfrac{\omega t}{2} $, then
\[
\overrightarrow{A}\cdot \overrightarrow{B} = \cos\Bigl(\omega t - \tfrac{\omega t}{2}\Bigr)
= \cos\Bigl(\tfrac{\omega t}{2}\Bigr).
\]
Step 5: Apply the Orthogonality Condition
For the vectors to be orthogonal:
\[
\overrightarrow{A}\cdot\overrightarrow{B} = 0
\quad \Longrightarrow \quad
\cos\Bigl(\tfrac{\omega t}{2}\Bigr) = 0.
\]
The general solutions to $ \cos(\theta) = 0 $ are $ \theta = \tfrac{\pi}{2} + n\pi $, where $ n $ is an integer.
Step 6: Solve for t
Setting $ \tfrac{\omega t}{2} = \tfrac{\pi}{2} $ (which gives the first positive time instant when they are perpendicular):
\[
\tfrac{\omega t}{2} = \tfrac{\pi}{2}
\quad \Longrightarrow \quad
t = \tfrac{\pi}{\omega}.
\]
Thus, the value of $ t $ at which $ \overrightarrow{A} $ and $ \overrightarrow{B} $ are orthogonal for the first time is
$ t = \tfrac{\pi}{\omega}. $
