The vectors `vecA=sin(alphat) hati-cos(alphat) hatj`and `vecB=cos(alphat^2/4) hati +sin(alphat^2/4) hatj` are orthogonal to each other the value of t would be(where `alpha` is a positive constant)

Options

A. t=2
B. t=4
C. t=6
D. t=8

Answer

To solve the problem of finding the value of \( t \) for which the vectors \( \vec{A} \) and \( \vec{B} \) are orthogonal, we will follow these steps:

### Step 1: Write down the vectors
The given vectors are:
\[ \vec{A} = \sin(\alpha t) \hat{i} - \cos(\alpha t) \hat{j} \]
\[ \vec{B} = \cos\left(\frac{\alpha t^2}{4}\right) \hat{i} + \sin\left(\frac{\alpha t^2}{4}\right) \hat{j} \]

### Step 2: Use the orthogonality condition
Two vectors are orthogonal if their dot product is zero:
\[ \vec{A} \cdot \vec{B} = 0 \]

### Step 3: Calculate the dot product
Calculating the dot product:
\[ \vec{A} \cdot \vec{B} = \left(\sin(\alpha t) \hat{i} - \cos(\alpha t) \hat{j}\right) \cdot \left(\cos\left(\frac{\alpha t^2}{4}\right) \hat{i} + \sin\left(\frac{\alpha t^2}{4}\right) \hat{j}\right) \]
Using the distributive property of the dot product:
\[ = \sin(\alpha t) \cos\left(\frac{\alpha t^2}{4}\right) + \left(-\cos(\alpha t)\right) \sin\left(\frac{\alpha t^2}{4}\right) \]
This simplifies to:
\[ \sin(\alpha t) \cos\left(\frac{\alpha t^2}{4}\right) - \cos(\alpha t) \sin\left(\frac{\alpha t^2}{4}\right) \]

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

### Step 5: Use the sine difference identity
Using the sine difference identity:
\[ \sin(a) \cos(b) - \cos(a) \sin(b) = \sin(a - b) \]
We can rewrite the equation as:
\[ \sin\left(\alpha t - \frac{\alpha t^2}{4}\right) = 0 \]

### Step 6: Solve for \( t \)
The sine function is zero when its argument is an integer multiple of \( \pi \):
\[ \alpha t - \frac{\alpha t^2}{4} = n\pi \quad (n \in \mathbb{Z}) \]
Rearranging gives:
\[ \alpha t - \frac{\alpha t^2}{4} = 0 \]
Factoring out \( \alpha t \):
\[ \alpha t \left(1 - \frac{t}{4}\right) = 0 \]
This gives two possible solutions:
1. \( \alpha t = 0 \) which implies \( t = 0 \)
2. \( 1 - \frac{t}{4} = 0 \) which implies \( t = 4 \)

### Step 7: Determine the valid solution
Since \( \alpha \) is a positive constant, \( t = 0 \) leads to both vectors being zero vectors, which is not valid. Therefore, the only valid solution is:
\[ t = 4 \]

### Final Answer
The value of \( t \) is:
\[ \boxed{4} \]