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`.

Explanation

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

### Step 1: Understand the Gravitational Field
The gravitational field vector \(\vec{E} = 3\hat{i} - 4\hat{j}\) indicates the direction and magnitude of the gravitational force per unit mass.

### Step 2: Rewrite the Line Equation
The equation of the line can be rewritten in slope-intercept form:
\[ y = \frac{3}{4}x + \frac{9}{4} \]
From this, we can identify the slope \(m_1 = \frac{3}{4}\).

### Step 3: Determine the Slope of the Gravitational Field
The gravitational field vector can be interpreted as a force vector. The slope of this vector can be calculated as:
\[ m_2 = \frac{\text{change in } y}{\text{change in } x} = \frac{-4}{3} \]

### Step 4: Calculate the Angle Between the Line and the Gravitational Field
To find the angle \(\theta\) between the line and the gravitational field, we can use the formula for the tangent of the angle between two slopes:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting the values:
\[ \tan \theta = \frac{\frac{3}{4} - \left(-\frac{4}{3}\right)}{1 + \left(\frac{3}{4}\right)\left(-\frac{4}{3}\right)} \]
Calculating the numerator:
\[ \frac{3}{4} + \frac{4}{3} = \frac{9}{12} + \frac{16}{12} = \frac{25}{12} \]
Calculating the denominator:
\[ 1 - 1 = 0 \]
Since the denominator is 0, this indicates that \(\tan \theta\) is undefined, which means \(\theta = 90^\circ\).

### Step 5: Calculate the Work Done
The work done \(W\) in a gravitational field when the displacement is perpendicular to the field is given by:
\[ W = F \cdot d \cdot \cos(\theta) \]
Since \(\theta = 90^\circ\), we have:
\[ \cos(90^\circ) = 0 \]
Thus, the work done is:
\[ W = F \cdot d \cdot 0 = 0 \text{ joules} \]

### Final Answer
The work done in displacing the particle by 1 meter along the given line in the gravitational field is \(0\) joules.

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