Find the torque `(vectau=vecrxxvecF)` of a force `vecF=-3hati+hatj+3hatk` acting at the point `vecr=7hati+3hatj+hatk`

Options

A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for statement-I
B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for statement-I
C. Statement-I is true, Statement-II is false.
D. Statement-I is false and Statement-II is true.

Answer

To find the torque \(\vec{\tau} = \vec{r} \times \vec{F}\) of a force \(\vec{F} = -3\hat{i} + \hat{j} + 3\hat{k}\) acting at the point \(\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}\), we will use the cross product formula.

### Step-by-Step Solution:

1. **Identify the Vectors**:
- The position vector \(\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}\)
- The force vector \(\vec{F} = -3\hat{i} + \hat{j} + 3\hat{k}\)

2. **Set Up the Determinant**:
We will use the determinant method to calculate the cross product. The determinant is set up as follows:

\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 3 \end{vmatrix} \]

3. **Calculate the Determinant**:
We will expand this determinant using the cofactor expansion:

\[ \vec{\tau} = \hat{i} \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 7 & 1 \\ -3 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 7 & 3 \\ -3 & 1 \end{vmatrix} \]

Now, we calculate each of the 2x2 determinants:

- For \(\hat{i}\):
\[ \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} = (3)(3) - (1)(1) = 9 - 1 = 8 \]

- For \(-\hat{j}\):
\[ \begin{vmatrix} 7 & 1 \\ -3 & 3 \end{vmatrix} = (7)(3) - (1)(-3) = 21 + 3 = 24 \]

- For \(\hat{k}\):
\[ \begin{vmatrix} 7 & 3 \\ -3 & 1 \end{vmatrix} = (7)(1) - (3)(-3) = 7 + 9 = 16 \]

4. **Combine the Results**:
Now substituting back into the expression for \(\vec{\tau}\):

\[ \vec{\tau} = 8\hat{i} - 24\hat{j} + 16\hat{k} \]

5. **Final Result**:
Thus, the torque vector is:

\[ \vec{\tau} = 8\hat{i} - 24\hat{j} + 16\hat{k} \]