The gravitational field in a region is given by `vecE = (3hati- 4hatj) N kg^(-1)`. Find out the work done (in joule) in displacing a particle by `1 m` along the line `4y = 3x + 9`.

Answer

To solve the problem of finding the work done in displacing a particle by 1 meter along the line given by \(4y = 3x + 9\) in a gravitational field represented by \(\vec{E} = (3\hat{i} - 4\hat{j}) \, \text{N kg}^{-1}\), we can follow these steps:

### Step 1: Understand the Gravitational Field
The gravitational field \(\vec{E}\) indicates the force per unit mass acting on a particle. The force \(\vec{F}\) acting on a particle of mass \(m\) in this field is given by:
\[ \vec{F} = m \vec{E} \]
Assuming \(m = 1 \, \text{kg}\), we have:
\[ \vec{F} = 1 \cdot (3\hat{i} - 4\hat{j}) = (3\hat{i} - 4\hat{j}) \, \text{N} \]

### Step 2: Determine the Displacement Vector
The line equation is given as \(4y = 3x + 9\). We can rewrite this as:
\[ y = \frac{3}{4}x + \frac{9}{4} \]
To find the displacement vector, we need two points on this line.

1. When \(x = 0\):
\[ y = \frac{9}{4} \quad \Rightarrow \quad (0, \frac{9}{4}) \]
2. When \(x = 1\):
\[ y = \frac{3}{4} \cdot 1 + \frac{9}{4} = \frac{12}{4} = 3 \quad \Rightarrow \quad (1, 3) \]

The displacement vector \(\vec{d}\) from point \((0, \frac{9}{4})\) to point \((1, 3)\) is:
\[ \vec{d} = (1 - 0)\hat{i} + \left(3 - \frac{9}{4}\right)\hat{j} = \hat{i} + \left(\frac{12}{4} - \frac{9}{4}\right)\hat{j} = \hat{i} + \frac{3}{4}\hat{j} \]

### Step 3: Calculate the Work Done
The work done \(W\) by the force \(\vec{F}\) during the displacement \(\vec{d}\) is given by the dot product:
\[ W = \vec{F} \cdot \vec{d} \]
Substituting the values:
\[ W = (3\hat{i} - 4\hat{j}) \cdot \left(\hat{i} + \frac{3}{4}\hat{j}\right) \]
Calculating the dot product:
\[ W = 3 \cdot 1 + (-4) \cdot \frac{3}{4} = 3 - 3 = 0 \]

### Conclusion
The work done in displacing the particle by 1 meter along the line is:
\[ \boxed{0 \, \text{J}} \]