A force`vecF=4hati-5hatj+3hatk`N is acting on a point `vecr_1=2hati+4hatj+3hatk`m. The torque acting about a point`vecr_2=4hati-3hatk`m is

Options

A. Zero
B. `42hati-30hatj+6hatk`Nm
C. `42hati+30hatj+6hatk`Nm`
D. `42hati+30hatj-6hatk`Nm

Answer

To find the torque acting about the point \(\vec{r_2}\) due to the force \(\vec{F}\) acting at the point \(\vec{r_1}\), we can use the formula for torque:

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

where \(\vec{r}\) is the position vector from the point about which we are calculating the torque (\(\vec{r_2}\)) to the point where the force is applied (\(\vec{r_1}\)), and \(\vec{F}\) is the force vector.

### Step 1: Calculate the position vector \(\vec{r}\)

The position vector \(\vec{r}\) is given by:

\[ \vec{r} = \vec{r_1} - \vec{r_2} \]

Substituting the values:

\[ \vec{r_1} = 2\hat{i} + 4\hat{j} + 3\hat{k} \]
\[ \vec{r_2} = 4\hat{i} - 3\hat{k} \]

So,

\[ \vec{r} = (2\hat{i} + 4\hat{j} + 3\hat{k}) - (4\hat{i} - 3\hat{k}) \]

Calculating this gives:

\[ \vec{r} = (2 - 4)\hat{i} + 4\hat{j} + (3 + 3)\hat{k} = -2\hat{i} + 4\hat{j} + 6\hat{k} \]

### Step 2: Write down the force vector \(\vec{F}\)

The force vector is given as:

\[ \vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k} \]

### Step 3: Calculate the torque \(\vec{\tau}\)

Now we can calculate the torque using the cross product:

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

Substituting the vectors:

\[ \vec{r} = -2\hat{i} + 4\hat{j} + 6\hat{k} \]
\[ \vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k} \]

We can set up the determinant for the cross product:

\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & 6 \\ 4 & -5 & 3 \end{vmatrix} \]

### Step 4: Calculate the determinant

Calculating the determinant:

\[ \vec{\tau} = \hat{i} \begin{vmatrix} 4 & 6 \\ -5 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 6 \\ 4 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 4 \\ 4 & -5 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):

\[ \begin{vmatrix} 4 & 6 \\ -5 & 3 \end{vmatrix} = (4)(3) - (6)(-5) = 12 + 30 = 42 \]

2. For \(\hat{j}\):

\[ \begin{vmatrix} -2 & 6 \\ 4 & 3 \end{vmatrix} = (-2)(3) - (6)(4) = -6 - 24 = -30 \]

3. For \(\hat{k}\):

\[ \begin{vmatrix} -2 & 4 \\ 4 & -5 \end{vmatrix} = (-2)(-5) - (4)(4) = 10 - 16 = -6 \]

Putting it all together:

\[ \vec{\tau} = 42\hat{i} + 30\hat{j} - 6\hat{k} \]

### Final Result

Thus, the torque acting about the point \(\vec{r_2}\) is:

\[ \vec{\tau} = 42\hat{i} + 30\hat{j} - 6\hat{k} \text{ Nm} \]