A force `vec(F)=3hat(i)+2hat(j)+chat(k)`N causes a displacement `vec(r )=chat(i)+4hat(j)+chat(k)` `m`. The work done is 36 `J`. Find the values `(s)` of c.

Explanation

To solve the problem step by step, we will follow the process of calculating the work done using the dot product of the force and displacement vectors.

### Step 1: Write down the given vectors
The force vector is given as:
\[ \vec{F} = 3\hat{i} + 2\hat{j} + c\hat{k} \, \text{N} \]
The displacement vector is given as:
\[ \vec{r} = c\hat{i} + 4\hat{j} + c\hat{k} \, \text{m} \]

### Step 2: Use the formula for work done
The work done \( W \) is given by the dot product of the force and displacement vectors:
\[ W = \vec{F} \cdot \vec{r} \]

### Step 3: Calculate the dot product
To find the dot product, we compute:
\[ \vec{F} \cdot \vec{r} = (3\hat{i} + 2\hat{j} + c\hat{k}) \cdot (c\hat{i} + 4\hat{j} + c\hat{k}) \]
This expands to:
\[ = 3c + 2 \cdot 4 + c \cdot c \]
\[ = 3c + 8 + c^2 \]

### Step 4: Set the equation for work done
We know from the problem statement that the work done \( W \) is 36 J. Therefore, we set up the equation:
\[ 3c + 8 + c^2 = 36 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ c^2 + 3c + 8 - 36 = 0 \]
\[ c^2 + 3c - 28 = 0 \]

### Step 6: Solve the quadratic equation
Now, we will solve the quadratic equation \( c^2 + 3c - 28 = 0 \) using the factorization method:
\[ c^2 + 7c - 4c - 28 = 0 \]
\[ (c + 7)(c - 4) = 0 \]

### Step 7: Find the values of \( c \)
Setting each factor to zero gives us:
1. \( c + 7 = 0 \) → \( c = -7 \)
2. \( c - 4 = 0 \) → \( c = 4 \)

Thus, the values of \( c \) are \( c = -7 \) and \( c = 4 \).

### Final Answer
The values of \( c \) are \( c = -7 \) and \( c = 4 \).

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