Dot product of two vectors `overset(rarr)A` and `overset(rarr)B` is defined as `overset(rarr)A.overset(rarr)B=aB cos phi` , where `phi` is angle between them when they are drawn with tails coinciding. For any two vectors . This means `overset(rarr)A . overset(rarr)B=overset(rarr)B. overset(rarr)A` that . The scalar product obeys the commutative law of multiplication, the order of the two vectors does not matter. The vector product of two vectors `overset(rarr)A` and `overset(rarr)B` also called the cross product, is denoted by `overset(rarr)A xx overset(rarr)B` . As the name suggests, the vector product is itself a vector. `overset(rarr)C=overset(rarr)A xx overset(rarr)B` then `C=AB sin theta` ,
A force `overset(rarr)F=3hat i +c hat j + 2 hatk` acting on a particle causes a displacement `d=4hat i- 2 hat j + 3 hat k` . If the work done (dot product of force and displacement) is 6J then the value of c is :

Options

A. 12
B. 0
C. 6
D. 1

Explanation

To solve the problem, we need to find the value of \( c \) in the force vector \( \vec{F} = 3 \hat{i} + c \hat{j} + 2 \hat{k} \) given that the work done by this force during a displacement \( \vec{d} = 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \) is 6 Joules.

### Step-by-step Solution:

1. **Write down the formula for work done:**
The work done \( W \) by a force \( \vec{F} \) over a displacement \( \vec{d} \) is given by the dot product:
\[ W = \vec{F} \cdot \vec{d} \]

2. **Substitute the given vectors into the work formula:**
We have:
\[ \vec{F} = 3 \hat{i} + c \hat{j} + 2 \hat{k} \]
\[ \vec{d} = 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \]
Therefore, the work done can be expressed as:
\[ W = (3 \hat{i} + c \hat{j} + 2 \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} + 3 \hat{k}) \]

3. **Calculate the dot product:**
The dot product is calculated by multiplying the corresponding components and summing them up:
\[ W = (3 \cdot 4) + (c \cdot -2) + (2 \cdot 3) \]
Simplifying this gives:
\[ W = 12 - 2c + 6 \]
Thus, we have:
\[ W = 18 - 2c \]

4. **Set the work done equal to 6 Joules:**
According to the problem, the work done is 6 Joules:
\[ 18 - 2c = 6 \]

5. **Solve for \( c \):**
Rearranging the equation:
\[ 18 - 6 = 2c \]
\[ 12 = 2c \]
Dividing both sides by 2:
\[ c = 6 \]

### Final Answer:
The value of \( c \) is \( 6 \).

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