Find the torque of a force `vecF=2hati+hatj+4hatk` acting at the point `vecr=7hati+3hatj+hatk` :

Options

A. `14hati-38hatj-16hatk`
B. `4hati+4hatj+6hatk`
C. `-14hati+38hatj-16hatk`
D. `11hati-26hatj+hatk`

Answer

To find the torque \(\vec{\tau}\) of a force \(\vec{F}\) acting at a point given by the position vector \(\vec{r}\), we use the formula:

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

where \(\times\) denotes the cross product.

**Step 1: Identify the vectors**
- The force vector is given as:
\[ \vec{F} = 2\hat{i} + \hat{j} + 4\hat{k} \]
- The position vector is given as:
\[ \vec{r} = 7\hat{i} + 3\hat{j} + \hat{k} \]

**Step 2: Set up the determinant for the cross product**
To find the cross product \(\vec{r} \times \vec{F}\), we can use the determinant method. We set up the determinant as follows:

\[ \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ 2 & 1 & 4 \end{vmatrix} \]

**Step 3: Calculate the determinant**
Calculating the determinant, we expand it as:

\[ \vec{r} \times \vec{F} = \hat{i} \begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 7 & 1 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 7 & 3 \\ 2 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix} = (3 \cdot 4) - (1 \cdot 1) = 12 - 1 = 11 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 7 & 1 \\ 2 & 4 \end{vmatrix} = (7 \cdot 4) - (1 \cdot 2) = 28 - 2 = 26 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 7 & 3 \\ 2 & 1 \end{vmatrix} = (7 \cdot 1) - (3 \cdot 2) = 7 - 6 = 1 \]

**Step 4: Combine the results**
Now, substituting back into the equation for \(\vec{r} \times \vec{F}\):

\[ \vec{r} \times \vec{F} = 11\hat{i} - 26\hat{j} + 1\hat{k} \]

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

\[ \vec{\tau} = 11\hat{i} - 26\hat{j} + 1\hat{k} \]

**Final Answer:**
The torque of the force is:
\[ \vec{\tau} = 11\hat{i} - 26\hat{j} + 1\hat{k} \]

---