If Vectors `vec(A)= cos omega t(i)+ sin omega t(j)` and `vec(B)=cosomegat/(2)hat(i)+sinomegat/(2)hat(j)`are functions of time. Then the value of `t` at which they are orthogonal to each other is

Options

A. `t = (pi)/(4 omega)`
B. `t = (pi)/(2 omega)`
C. `t = (pi)/(omega)`
D. `t=0`

Answer

To determine the value of \( t \) at which the vectors \( \vec{A} \) and \( \vec{B} \) are orthogonal, we need to follow these steps:

### Step 1: Understand the condition for orthogonality
Two vectors are orthogonal if their dot product is zero. Therefore, we need to find when \( \vec{A} \cdot \vec{B} = 0 \).

### Step 2: Write the expressions for the vectors
Given:
\[ \vec{A} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \]
\[ \vec{B} = \cos\left(\frac{\omega t}{2}\right) \hat{i} + \sin\left(\frac{\omega t}{2}\right) \hat{j} \]

### Step 3: Compute the dot product
The dot product \( \vec{A} \cdot \vec{B} \) is calculated as follows:
\[ \vec{A} \cdot \vec{B} = \left(\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}\right) \cdot \left(\cos\left(\frac{\omega t}{2}\right) \hat{i} + \sin\left(\frac{\omega t}{2}\right) \hat{j}\right) \]
Using the properties of dot products:
\[ \vec{A} \cdot \vec{B} = \cos(\omega t) \cos\left(\frac{\omega t}{2}\right) + \sin(\omega t) \sin\left(\frac{\omega t}{2}\right) \]

### Step 4: Set the dot product to zero
Setting the dot product equal to zero for orthogonality:
\[ \cos(\omega t) \cos\left(\frac{\omega t}{2}\right) + \sin(\omega t) \sin\left(\frac{\omega t}{2}\right) = 0 \]

### Step 5: Use the cosine addition formula
Using the cosine addition formula, we can rewrite the equation:
\[ \cos\left(\omega t - \frac{\omega t}{2}\right) = 0 \]
This simplifies to:
\[ \cos\left(\frac{\omega t}{2}\right) = 0 \]

### Step 6: Solve for \( t \)
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ \frac{\omega t}{2} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Multiplying through by 2 gives:
\[ \omega t = \pi + 2n\pi \]
Thus,
\[ t = \frac{\pi(1 + 2n)}{\omega} \]

### Step 7: Find the first positive solution
For the smallest positive \( t \), set \( n = 0 \):
\[ t = \frac{\pi}{\omega} \]

### Final Answer
The value of \( t \) at which the vectors \( \vec{A} \) and \( \vec{B} \) are orthogonal is:
\[ t = \frac{\pi}{\omega} \]