The torque of force `F = -3 hat(i)+hat(j) + 5 hat(k)` acting on a point `r = 7 hat(i) + 3 hat(j) + hat(k)` about origin will be

Options

A. `14 hat(i)-38 hat(j) + 16 hat(k)`
B. `4 hat(i)+4hat(j)+6 hat(k)`
C. `-14 hat(i)+38 hat(j)-16 hat(k)`
D. `-21 hat(i)+3 hat(j)+5 hat(k)`

Explanation

To find the torque \( \tau \) of the force \( \mathbf{F} \) acting on a point \( \mathbf{r} \) about the origin, we use the formula:

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

### Step 1: Write down the vectors
Given:
\[ \mathbf{F} = -3 \hat{i} + \hat{j} + 5 \hat{k} \]
\[ \mathbf{r} = 7 \hat{i} + 3 \hat{j} + \hat{k} \]

### Step 2: Set up the cross product
We need to compute the cross product \( \mathbf{r} \times \mathbf{F} \). We can set this up using a determinant:

\[ \tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we expand it as follows:

\[ \tau = \hat{i} \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} 7 & 1 \\ -3 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} 7 & 3 \\ -3 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} = (3 \cdot 5) - (1 \cdot 1) = 15 - 1 = 14 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 7 & 1 \\ -3 & 5 \end{vmatrix} = (7 \cdot 5) - (1 \cdot -3) = 35 + 3 = 38 \]

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

### Step 4: Combine the results
Now substituting back into the expression for \( \tau \):

\[ \tau = 14 \hat{i} - 38 \hat{j} + 16 \hat{k} \]

### Final Result
Thus, the torque \( \tau \) is:

\[ \tau = 14 \hat{i} - 38 \hat{j} + 16 \hat{k} \]