If the velocity of a particle is `(2hati+3hatj-4hatk)` and its acceleration is `(-hati+2hatj+hatk)` and angle between them in `(npi)/(4)`. The value of n is .

Answer

To solve the problem, we need to find the value of \( n \) given the velocity vector \( \mathbf{v} = 2\hat{i} + 3\hat{j} - 4\hat{k} \) and the acceleration vector \( \mathbf{a} = -\hat{i} + 2\hat{j} + \hat{k} \), with the angle between them expressed as \( \theta = \frac{n\pi}{4} \).

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Velocity vector: \( \mathbf{v} = 2\hat{i} + 3\hat{j} - 4\hat{k} \)
- Acceleration vector: \( \mathbf{a} = -\hat{i} + 2\hat{j} + \hat{k} \)

2. **Calculate the Dot Product**:
The dot product of two vectors \( \mathbf{v} \) and \( \mathbf{a} \) is given by:
\[ \mathbf{v} \cdot \mathbf{a} = (2)(-1) + (3)(2) + (-4)(1) \]
Simplifying this:
\[ \mathbf{v} \cdot \mathbf{a} = -2 + 6 - 4 = 0 \]

3. **Calculate the Magnitudes of the Vectors**:
- Magnitude of \( \mathbf{v} \):
\[ |\mathbf{v}| = \sqrt{(2)^2 + (3)^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]
- Magnitude of \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{(-1)^2 + (2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

4. **Use the Dot Product Formula**:
The dot product can also be expressed in terms of the magnitudes and the cosine of the angle between them:
\[ \mathbf{v} \cdot \mathbf{a} = |\mathbf{v}| |\mathbf{a}| \cos(\theta) \]
Substituting the values we calculated:
\[ 0 = \sqrt{29} \cdot \sqrt{6} \cdot \cos(\theta) \]

5. **Determine the Angle**:
Since the dot product is zero, it implies:
\[ \cos(\theta) = 0 \]
The angle \( \theta \) for which cosine is zero is:
\[ \theta = \frac{\pi}{2} \text{ radians} \]

6. **Relate the Angle to \( n \)**:
According to the problem, we have:
\[ \theta = \frac{n\pi}{4} \]
Setting this equal to \( \frac{\pi}{2} \):
\[ \frac{n\pi}{4} = \frac{\pi}{2} \]

7. **Solve for \( n \)**:
To find \( n \), we can cross-multiply:
\[ n\pi = 2\pi \quad \Rightarrow \quad n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).