Under a force of `10 hati - 3 hatj +6 hatk` newton, a body of mass 5 kg is displaced from the position `6 hati + 5 hatj - 3 hatk` to the position `10 hati - 2hatj + 7 hatk`. Calculate the work done.

Answer

To calculate the work done by the force on the body during its displacement, we can follow these steps:

### Step 1: Identify the Force and the Displacement Vectors
The force vector \( \mathbf{F} \) is given as:
\[ \mathbf{F} = 10 \hat{i} - 3 \hat{j} + 6 \hat{k} \text{ N} \]

The initial position vector \( \mathbf{r_1} \) is:
\[ \mathbf{r_1} = 6 \hat{i} + 5 \hat{j} - 3 \hat{k} \]

The final position vector \( \mathbf{r_2} \) is:
\[ \mathbf{r_2} = 10 \hat{i} - 2 \hat{j} + 7 \hat{k} \]

### Step 2: Calculate the Displacement Vector
The displacement vector \( \Delta \mathbf{r} \) is calculated as:
\[ \Delta \mathbf{r} = \mathbf{r_2} - \mathbf{r_1} \]
Calculating each component:
- For \( \hat{i} \): \( 10 - 6 = 4 \)
- For \( \hat{j} \): \( -2 - 5 = -7 \)
- For \( \hat{k} \): \( 7 - (-3) = 10 \)

Thus, the displacement vector is:
\[ \Delta \mathbf{r} = 4 \hat{i} - 7 \hat{j} + 10 \hat{k} \]

### Step 3: Calculate the Work Done
The work done \( W \) by the force is given by the dot product of the force vector and the displacement vector:
\[ W = \mathbf{F} \cdot \Delta \mathbf{r} \]
Calculating the dot product:
\[ W = (10 \hat{i} - 3 \hat{j} + 6 \hat{k}) \cdot (4 \hat{i} - 7 \hat{j} + 10 \hat{k}) \]
Calculating each term:
- \( 10 \times 4 = 40 \)
- \( -3 \times -7 = 21 \)
- \( 6 \times 10 = 60 \)

Adding these values together:
\[ W = 40 + 21 + 60 = 121 \text{ J} \]

### Final Answer
The work done by the force is:
\[ W = 121 \text{ J} \]