Find the torque of a force `(7 hati + 3hatj - 5hatk)` about the origin. The force acts on a particle whose position vector is `(hati - hatj + hatk)`.

Options

A. `2hati+12hatj+10hatk`
B. `2hati+10hatj+12hatk`
C. `2hati+10hatj+10hatk`
D. `10hati+2hatj+hatk`

Explanation

To find the torque of a force about the origin, we can use the formula for torque, which is defined as the cross product of the position vector (R) and the force vector (F):

\[ \text{Torque} (\tau) = \mathbf{R} \times \mathbf{F} \]

### Step 1: Identify the position vector and force vector
The position vector \(\mathbf{R}\) is given as:
\[ \mathbf{R} = \hat{i} - \hat{j} + \hat{k} \]
The force vector \(\mathbf{F}\) is given as:
\[ \mathbf{F} = 7\hat{i} + 3\hat{j} - 5\hat{k} \]

### Step 2: Set up the cross product
We will calculate the cross product \(\mathbf{R} \times \mathbf{F}\) using the determinant of a matrix formed by the unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\), and the components of the vectors \(\mathbf{R}\) and \(\mathbf{F}\).

\[ \mathbf{R} \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we can expand it using the first row:

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

### Step 4: Calculate the 2x2 determinants
1. For \(\hat{i}\):
\[ \begin{vmatrix} -1 & 1 \\ 3 & -5 \end{vmatrix} = (-1)(-5) - (1)(3) = 5 - 3 = 2 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 1 & 1 \\ 7 & -5 \end{vmatrix} = (1)(-5) - (1)(7) = -5 - 7 = -12 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 1 & -1 \\ 7 & 3 \end{vmatrix} = (1)(3) - (-1)(7) = 3 + 7 = 10 \]

### Step 5: Substitute back into the equation
Now substituting back into the expression for torque:

\[ \mathbf{R} \times \mathbf{F} = 2\hat{i} - (-12)\hat{j} + 10\hat{k} = 2\hat{i} + 12\hat{j} + 10\hat{k} \]

### Final Result
Thus, the torque of the force about the origin is:
\[ \mathbf{\tau} = 2\hat{i} + 12\hat{j} + 10\hat{k} \]