Question

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

• A. t = $ \frac{ \pi}{ \omega } $
• B. t = 0
• C. t = $ \frac{ \pi}{ 4 \omega } $
• D. t = $ \frac{ \pi}{2 \omega } $

Answer 1

The Correct Option is A

To determine the value of $t$ at which the vectors $ \overrightarrow{A} = \cos \, \omega t \, \widehat{i} + \sin \, \omega \, t \, \widehat{j} $ and $ \overrightarrow{B} = \cos \, \frac{\omega \, t}{2} \, \widehat{i} + \sin \, \frac{\omega \, t}{2} \, \widehat{j} $ are orthogonal to each other, we need to evaluate when their dot product is zero.

The dot product of two vectors $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is given by:

$$ \overrightarrow{A} \cdot \overrightarrow{B} = (\cos \omega t )(\cos \frac{\omega t}{2}) + (\sin \omega t )(\sin \frac{\omega t}{2}) $$

For the vectors to be orthogonal, the dot product must be zero:

$$ \cos \omega t \cdot \cos \frac{\omega t}{2} + \sin \omega t \cdot \sin \frac{\omega t}{2} = 0 $$

Using the trigonometric identity for cosine of the sum of angles:

$$ \cos(A - B) = \cos A \cos B + \sin A \sin B $$

We can rewrite the equation as:

$$ \cos \left(\omega t - \frac{\omega t}{2}\right) = \cos \left(\frac{\omega t}{2}\right) = 0 $$

The cosine function equals zero at odd multiples of $\frac{\pi}{2}$, hence:

$$ \frac{\omega t}{2} = (2n+1)\frac{\pi}{2} $$

Solving for $t$ when $n = 0$:

$$ \frac{\omega t}{2} = \frac{\pi}{2} $$

$$ \omega t = \pi $$

$$ t = \frac{\pi}{\omega} $$

Therefore, the value of $t$ at which the vectors $ \overrightarrow{A} $ and $ \overrightarrow{B} $ are orthogonal to each other is $ \frac{\pi}{\omega} $.