The force `7 hat i+ 3 hat j - 5 hat k` acts on a particle whose position vector is `hat i- hat j + hat k`. What is the torque of a given force about the origin ?

Options

A. `2 hat i+ 12 hat j + 10 hat k`
B. `2 hat i+ 10 hat j + 12 hat k`
C. `2 hat i+10 hat j + 10 hat k`
D. `10 hat i+ 2 hat j + hat k`

Explanation

To find the torque of a given force about the origin, we can use the formula for torque \(\vec{\tau}\):

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

where:
- \(\vec{\tau}\) is the torque,
- \(\vec{r}\) is the position vector,
- \(\vec{F}\) is the force vector,
- \(\times\) denotes the cross product.

### Step 1: Identify the vectors
Given:
- Force vector \(\vec{F} = 7\hat{i} + 3\hat{j} - 5\hat{k}\)
- Position vector \(\vec{r} = \hat{i} - \hat{j} + \hat{k}\)

### Step 2: Set up the cross product
We need to compute the cross product \(\vec{r} \times \vec{F}\):

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

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

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

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} -1 & 1 \\ 3 & -5 \end{vmatrix} = (-1)(-5) - (1)(3) = 5 - 3 = 2 \]

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

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

### Step 4: Combine the results
Now, substituting back into the torque equation:

\[ \vec{\tau} = 2\hat{i} + 12\hat{j} + 10\hat{k} \]

### Final Result
Thus, the torque of the given force about the origin is:

\[ \vec{\tau} = 2\hat{i} + 12\hat{j} + 10\hat{k} \]