Find the torque of a force `vecF= -3hati+hatj+5hatk ` acting at the point `vecr=7hati+3hatj+hatk`

Options

A. `-21hati+3hatj+5hatk`
B. `-14hati+3hatj-16hatk`
C. `4hati+4hati+6hatk`
D. `14hati-38hatj+16hatk`

Answer

To find the torque of the force \(\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}\) acting at the point given by the position vector \(\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}\), we can use the formula for torque:

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

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

### Step 2: Set up the cross product
The torque \(\vec{\tau}\) can be calculated using the determinant of a matrix formed by the unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\), and the components of \(\vec{r}\) and \(\vec{F}\):

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

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

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

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 \(\vec{\tau}\):

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

### Final Answer
Thus, the torque \(\vec{\tau}\) is:

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