A force `vec(F)=(2hat(i)+3hat(j)-5hat(k))N` acts at a point `vec(r )_(1)=(2hat(i)+4hat(j)+7hat(k))m`. The torque of the force about the point `vec(r )_(2)=(hat(i)+2hat(j)+3hat(k))m` is

Options

A. `(17hat(j)+5hat(k)-3hat(i))Nm`
B. `(2hat(j)+4hat(j)-6hat(k))Nm`
C. `(12hat(i)-5hat(j)+7hat(k))Nm`
D. `-(-13hat(j)+22hat(i)+hat(k))Nm`

Answer

To find the torque of the force \(\vec{F} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \, \text{N}\) about the point \(\vec{r}_2 = (\hat{i} + 2\hat{j} + 3\hat{k}) \, \text{m}\), we will follow these steps:

### Step 1: Determine the position vector \(\vec{r}\)
The torque \(\vec{\tau}\) is given by the equation:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
where \(\vec{r}\) is the position vector from the point \(\vec{r}_2\) to the point \(\vec{r}_1\) where the force is applied.

First, we calculate \(\vec{r}\):
\[ \vec{r} = \vec{r}_1 - \vec{r}_2 \]
Given:
\[ \vec{r}_1 = (2\hat{i} + 4\hat{j} + 7\hat{k}) \, \text{m} \]
\[ \vec{r}_2 = (\hat{i} + 2\hat{j} + 3\hat{k}) \, \text{m} \]
Now, subtract \(\vec{r}_2\) from \(\vec{r}_1\):
\[ \vec{r} = (2\hat{i} + 4\hat{j} + 7\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (2 - 1)\hat{i} + (4 - 2)\hat{j} + (7 - 3)\hat{k} \]
\[ \vec{r} = (1\hat{i} + 2\hat{j} + 4\hat{k}) \, \text{m} \]

### Step 2: Calculate the torque \(\vec{\tau}\)
Now we can calculate the torque using the cross product:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
Substituting the values:
\[ \vec{F} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \, \text{N} \]
We set up the determinant for the cross product:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 3 & -5 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} 2 & 4 \\ 3 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 4 \\ 2 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} 2 & 4 \\ 3 & -5 \end{vmatrix} = (2)(-5) - (4)(3) = -10 - 12 = -22 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 1 & 4 \\ 2 & -5 \end{vmatrix} = (1)(-5) - (4)(2) = -5 - 8 = -13 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1 \]

Putting it all together:
\[ \vec{\tau} = -22\hat{i} + 13\hat{j} - 1\hat{k} \]

### Step 4: Final expression for torque
Thus, the torque is:
\[ \vec{\tau} = -22\hat{i} + 13\hat{j} - 1\hat{k} \]