velocity and acceleration of a particle at some instant are
`v=(3 hati-4 hatj+2 hatk) m//s and a=(2 hati+hatj-2hatk) m//s^2`
(a) What is the value of dot product of v and a at the given instant?
(b) What is the angle between v and a, acute, obtuse or `90^@`?
(c) At the given instant, whether speed of the particle is increasing, decreasing or constant?

Explanation

To solve the problem step by step, we will break it down into three parts as per the question.

### Given:
- Velocity vector, \( \mathbf{v} = 3 \hat{i} - 4 \hat{j} + 2 \hat{k} \) m/s
- Acceleration vector, \( \mathbf{a} = 2 \hat{i} + \hat{j} - 2 \hat{k} \) m/s²

### (a) Calculate the dot product of \( \mathbf{v} \) and \( \mathbf{a} \)

The dot product of two vectors \( \mathbf{v} \) and \( \mathbf{a} \) is given by the formula:
\[ \mathbf{v} \cdot \mathbf{a} = v_x a_x + v_y a_y + v_z a_z \]

Where:
- \( v_x = 3 \), \( v_y = -4 \), \( v_z = 2 \)
- \( a_x = 2 \), \( a_y = 1 \), \( a_z = -2 \)

Substituting these values into the formula:
\[ \mathbf{v} \cdot \mathbf{a} = (3)(2) + (-4)(1) + (2)(-2) \]
\[ = 6 - 4 - 4 \]
\[ = -2 \]

**Dot Product Result:**
\[ \mathbf{v} \cdot \mathbf{a} = -2 \]

### (b) Determine the angle between \( \mathbf{v} \) and \( \mathbf{a} \)

The dot product can also be expressed in terms of the angle \( \theta \) between the two vectors:
\[ \mathbf{v} \cdot \mathbf{a} = |\mathbf{v}| |\mathbf{a}| \cos \theta \]

Since we found \( \mathbf{v} \cdot \mathbf{a} = -2 \), and knowing that the magnitudes of \( \mathbf{v} \) and \( \mathbf{a} \) are always positive, we can conclude that:
- If \( \cos \theta < 0 \), then \( \theta \) is obtuse (greater than \( 90^\circ \)).

Since the dot product is negative, the angle between \( \mathbf{v} \) and \( \mathbf{a} \) is obtuse.

**Angle Result:**
The angle between \( \mathbf{v} \) and \( \mathbf{a} \) is obtuse.

### (c) Determine if the speed of the particle is increasing, decreasing, or constant

The relationship between velocity and acceleration can help us determine the behavior of speed:
- If the angle between velocity and acceleration is acute (\( < 90^\circ \)), speed is increasing.
- If the angle is obtuse (\( > 90^\circ \)), speed is decreasing.
- If the angle is \( 90^\circ \), speed is constant.

Since we established that the angle between \( \mathbf{v} \) and \( \mathbf{a} \) is obtuse, it follows that the speed of the particle is decreasing.

**Speed Result:**
At the given instant, the speed of the particle is decreasing.

### Summary of Results:
(a) Dot Product \( \mathbf{v} \cdot \mathbf{a} = -2 \)
(b) Angle between \( \mathbf{v} \) and \( \mathbf{a} \) is obtuse
(c) Speed of the particle is decreasing