The torque of a force `F = - 6 hat (i)` acting at a point `r = 4 hat(j)` about origin will be

Options

A. `-24 hat(k)`
B. `24 hat(k)`
C. `24 hat(j)`
D. `24 hat(i)`

Explanation

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

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

Where:
- \( \mathbf{r} \) is the position vector,
- \( \mathbf{F} \) is the force vector,
- \( \times \) denotes the cross product.

### Step 1: Identify the vectors
Given:
- \( \mathbf{F} = -6 \hat{i} \)
- \( \mathbf{r} = 4 \hat{j} \)

### Step 2: Write the vectors in component form
The vectors can be expressed as:
- \( \mathbf{F} = 0 \hat{i} + 0 \hat{j} - 6 \hat{k} \)
- \( \mathbf{r} = 0 \hat{i} + 4 \hat{j} + 0 \hat{k} \)

### Step 3: Calculate the cross product
To find the torque, we calculate \( \mathbf{r} \times \mathbf{F} \):

\[ \tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 0 \\ -6 & 0 & 0 \end{vmatrix} \]

### Step 4: Evaluate the determinant
Using the determinant, we can calculate:

\[ \tau = \hat{i} \begin{vmatrix} 4 & 0 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 0 \\ -6 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 4 \\ -6 & 0 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 4 & 0 \\ 0 & 0 \end{vmatrix} = (4)(0) - (0)(0) = 0 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 0 & 0 \\ -6 & 0 \end{vmatrix} = (0)(0) - (0)(-6) = 0 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 0 & 4 \\ -6 & 0 \end{vmatrix} = (0)(0) - (4)(-6) = 24 \]

### Step 5: Combine the results
Putting it all together, we have:

\[ \tau = 0 \hat{i} - 0 \hat{j} + 24 \hat{k} = 24 \hat{k} \]

### Final Answer
Thus, the torque \( \tau \) about the origin is:

\[ \tau = 24 \hat{k} \]

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