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

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

where \( \vec{r} = \vec{r_2} - \vec{r_1} \).

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

Given:
- \( \vec{r_1} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k} \) m
- \( \vec{r_2} = 4 \hat{i} - 3 \hat{k} \) m

We find \( \vec{r} \) as follows:

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

Calculating this gives:

\[ \vec{r} = (4 - 2) \hat{i} + (0 - 4) \hat{j} + (-3 - 3) \hat{k} = 2 \hat{i} - 4 \hat{j} - 6 \hat{k} \]

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

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

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

Using the cross product \( \vec{\tau} = \vec{r} \times \vec{F} \):

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

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 \( \hat{i} \):
\[ (-4)(3) - (-6)(-5) = -12 - 30 = -42 \]

2. For \( \hat{j} \):
\[ (2)(3) - (-6)(4) = 6 + 24 = 30 \]

3. For \( \hat{k} \):
\[ (2)(-5) - (-4)(4) = -10 + 16 = 6 \]

Putting it all together:

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

### Final Result

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

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