A force `vec(F)=4hati-5hatj+3hatk` is acting a point `vec(r_(1))=hati+2hatj+3hatk`. The torque acting about a point `vec(r_(2))=3hati-2hatj-3hatk` is

Options

A. zero
B. `42hati-30hatj+6hatk`
C. `42hati+30hatj+6hatk`
D. `42hati+30hatj-6hatk`

Answer

To find the torque acting about a point due to a force, we can use the formula:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

where \(\vec{\tau}\) is the torque, \(\vec{r}\) is the position vector from the point about which we are calculating the torque to the point where the force is applied, and \(\vec{F}\) is the force vector.

### Step-by-Step Solution:

1. **Identify the vectors:**
- The force vector is given as:
\[ \vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k} \]
- The position vector where the force is applied is:
\[ \vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k} \]
- The point about which we are calculating the torque is:
\[ \vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k} \]

2. **Calculate the position vector \(\vec{r}\) from \(\vec{r_2}\) to \(\vec{r_1}\):**
- We find \(\vec{r}\) by subtracting \(\vec{r_2}\) from \(\vec{r_1}\):
\[ \vec{r} = \vec{r_1} - \vec{r_2} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (3\hat{i} - 2\hat{j} - 3\hat{k}) \]
- Simplifying this gives:
\[ \vec{r} = \hat{i} - 3\hat{i} + 2\hat{j} + 2\hat{j} + 3\hat{k} + 3\hat{k} = -2\hat{i} + 4\hat{j} + 6\hat{k} \]

3. **Calculate the torque \(\vec{\tau}\):**
- Now we can calculate the torque using the cross product:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
- Setting up the determinant for the cross product:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & 6 \\ 4 & -5 & 3 \end{vmatrix} \]

4. **Calculate the determinant:**
- Expanding the determinant:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} 4 & 6 \\ -5 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 6 \\ 4 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 4 \\ 4 & -5 \end{vmatrix} \]
- Calculating each of the 2x2 determinants:
- For \(\hat{i}\):
\[ 4 \cdot 3 - (-5) \cdot 6 = 12 + 30 = 42 \]
- For \(\hat{j}\):
\[ -2 \cdot 3 - 4 \cdot 6 = -6 - 24 = -30 \quad \text{(but we take -(-30) = +30)} \]
- For \(\hat{k}\):
\[ -2 \cdot (-5) - 4 \cdot 4 = 10 - 16 = -6 \]

5. **Combine the results:**
- Thus, we have:
\[ \vec{\tau} = 42\hat{i} + 30\hat{j} - 6\hat{k} \]

### Final Answer:
The torque acting about the point \(\vec{r_2}\) is:
\[ \vec{\tau} = 42\hat{i} + 30\hat{j} - 6\hat{k} \]