Let `(veca (x) = (sin x) hati+ (cos x) hatj and vecb(x) = (cos 2x) hati + (sin 2x) hatj` be two variable vectors `( x in R)`. Then `veca (x) and vecb (x) ` are

Options

A. collinear for unique value of x
B. perpendicular for infinte values of x.
C. zero vectors for unique value of x
D. none of these

Explanation

To determine the relationship between the vectors \(\vec{a}(x) = \sin x \hat{i} + \cos x \hat{j}\) and \(\vec{b}(x) = \cos 2x \hat{i} + \sin 2x \hat{j}\), we will analyze if they are perpendicular.

### Step-by-step Solution:

1. **Understanding the Vectors**:
- We have two vectors defined as:
\[ \vec{a}(x) = \sin x \hat{i} + \cos x \hat{j} \]
\[ \vec{b}(x) = \cos 2x \hat{i} + \sin 2x \hat{j} \]

2. **Condition for Perpendicular Vectors**:
- Two vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular if their dot product is zero:
\[ \vec{a}(x) \cdot \vec{b}(x) = 0 \]

3. **Calculating the Dot Product**:
- The dot product of \(\vec{a}(x)\) and \(\vec{b}(x)\) is given by:
\[ \vec{a}(x) \cdot \vec{b}(x) = (\sin x)(\cos 2x) + (\cos x)(\sin 2x) \]

4. **Using the Sine Addition Formula**:
- We can use the sine addition formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
- Here, let \(A = x\) and \(B = 2x\):
\[ \sin(x + 2x) = \sin(3x) \]
- Thus, we have:
\[ \sin 3x = \sin x \cos 2x + \cos x \sin 2x \]

5. **Setting the Dot Product to Zero**:
- For the vectors to be perpendicular, we set the dot product to zero:
\[ \sin 3x = 0 \]

6. **Finding Solutions**:
- The solutions to \(\sin 3x = 0\) occur when:
\[ 3x = n\pi \quad \text{for } n \in \mathbb{Z} \]
- Therefore, we can solve for \(x\):
\[ x = \frac{n\pi}{3} \]

7. **Conclusion**:
- The vectors \(\vec{a}(x)\) and \(\vec{b}(x)\) are perpendicular for infinitely many values of \(x\) given by \(x = \frac{n\pi}{3}\) where \(n\) is any integer.

### Final Answer:
The vectors \(\vec{a}(x)\) and \(\vec{b}(x)\) are perpendicular for infinitely many values of \(x\).