A particle is projected in such a way that it follows a curved path with constant acceleration `vec(a)`. For finite interval of motion. Which of the following option `(s)` may be correct `:`
`vec(u)=` initial velocity `vec(a)=` acceleration of particle `vec(v)=` velocity at `tgt0`

Options

A. `|vec(a)xxvec(u)|cancel(=)0`
B. `|vec(a)xxvec(v)|=0`
C. `|vec(u)xxvec(v)|=0`
D. `|vec(u).vec(v)|=0`

Explanation

To solve the problem, we need to analyze the motion of a particle projected in a curved path with constant acceleration. We will evaluate the four options given in the question, focusing on the relationships between initial velocity (u), acceleration (a), and velocity at time t (v).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is projected with an initial velocity \( \vec{u} \).
- It experiences a constant acceleration \( \vec{a} \).
- The velocity at time \( t \) is given by \( \vec{v} = \vec{u} + \vec{a} t \).

2. **Analyzing the First Option**:
- The first option states that \( \vec{a} \times \vec{u} \neq 0 \).
- The cross product \( \vec{a} \times \vec{u} \) is zero only if \( \vec{a} \) and \( \vec{u} \) are parallel (i.e., the angle between them is 0 degrees or 180 degrees).
- Since the particle is moving in a curved path, \( \vec{a} \) cannot be parallel to \( \vec{u} \) for the entire motion. Therefore, this option is **correct**.

3. **Analyzing the Second Option**:
- The second option states that \( \vec{a} \times \vec{v} = 0 \).
- Since \( \vec{v} = \vec{u} + \vec{a} t \), we can express the cross product as \( \vec{a} \times (\vec{u} + \vec{a} t) \).
- The cross product of \( \vec{a} \) with itself is zero, but \( \vec{a} \times \vec{u} \) is not zero (as established in the first option). Thus, this option is **incorrect**.

4. **Analyzing the Third Option**:
- The third option states that \( \vec{u} \times \vec{v} = 0 \).
- Since \( \vec{v} = \vec{u} + \vec{a} t \), we have \( \vec{u} \times \vec{v} = \vec{u} \times (\vec{u} + \vec{a} t) \).
- The cross product \( \vec{u} \times \vec{u} = 0 \), so this option is also **incorrect**.

5. **Analyzing the Fourth Option**:
- The fourth option states that \( \vec{u} \cdot \vec{v} = 0 \).
- The dot product \( \vec{u} \cdot \vec{v} = \vec{u} \cdot (\vec{u} + \vec{a} t) = \vec{u} \cdot \vec{u} + \vec{u} \cdot \vec{a} t \).
- This dot product is zero only if the particle is in circular motion, where \( \vec{u} \) and \( \vec{a} \) are perpendicular. Thus, this option is **correct** for circular motion.

### Final Evaluation of Options:
- **Option 1**: Correct (\( \vec{a} \times \vec{u} \neq 0 \))
- **Option 2**: Incorrect (\( \vec{a} \times \vec{v} \neq 0 \))
- **Option 3**: Incorrect (\( \vec{u} \times \vec{v} \neq 0 \))
- **Option 4**: Correct (\( \vec{u} \cdot \vec{v} = 0 \) for circular motion)

### Summary of Correct Options:
The correct options are:
- Option 1: \( \vec{a} \times \vec{u} \neq 0 \)
- Option 4: \( \vec{u} \cdot \vec{v} = 0 \) (for circular motion)