A block is displaced from `(1m, 4m, 6m)` to `(2hati+3hatj-4hatk) m` under a constant force `F=(6hati-2hatj + hatk) N`. Find the work done by this force.

Explanation

To find the work done by the force on the block, we can follow these steps:

### Step 1: Identify the Initial and Final Positions
The initial position of the block is given as \( (1, 4, 6) \) m, which can be represented in vector form as:
\[ \mathbf{r}_{\text{initial}} = 1 \hat{i} + 4 \hat{j} + 6 \hat{k} \]
The final position is given as \( (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) \) m, which can be represented as:
\[ \mathbf{r}_{\text{final}} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k} \]

### Step 2: Calculate the Displacement Vector
The displacement vector \( \mathbf{d} \) can be calculated as:
\[ \mathbf{d} = \mathbf{r}_{\text{final}} - \mathbf{r}_{\text{initial}} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) - (1 \hat{i} + 4 \hat{j} + 6 \hat{k}) \]
Calculating this gives:
\[ \mathbf{d} = (2 - 1) \hat{i} + (3 - 4) \hat{j} + (-4 - 6) \hat{k} = 1 \hat{i} - 1 \hat{j} - 10 \hat{k} \]

### Step 3: Identify the Force Vector
The force vector \( \mathbf{F} \) is given as:
\[ \mathbf{F} = 6 \hat{i} - 2 \hat{j} + 1 \hat{k} \]

### Step 4: Calculate the Work Done
The work done \( W \) by the force can be calculated using the dot product of the force vector and the displacement vector:
\[ W = \mathbf{F} \cdot \mathbf{d} \]
Calculating the dot product:
\[ W = (6 \hat{i} - 2 \hat{j} + 1 \hat{k}) \cdot (1 \hat{i} - 1 \hat{j} - 10 \hat{k}) \]
Using the dot product formula:
\[ W = (6 \cdot 1) + (-2 \cdot -1) + (1 \cdot -10) \]
Calculating each term:
\[ W = 6 + 2 - 10 = -2 \text{ Joules} \]

### Final Answer
The work done by the force is:
\[ W = -2 \text{ Joules} \]

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