The torque of a force `F = 2 hat(i) - 3 hat(j) +5 hat(k)` acting at a point whose position vector `r = 3 hat(i) - 3 hat(j) +5 hat(k)` about the origin is

Options

A. `-3 hat(i)+5hat(k)`
B. `-5hat(i)+3hat(k)`
C. `-5hat(j)-3hat(k)`
D. `3 hat(i)-5hat(j)+3hat(k)`

Explanation

To find the torque \(\vec{T}\) of a force \(\vec{F}\) acting at a point with position vector \(\vec{r}\) about the origin, we use the formula:

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

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

### Step 1: Set up the cross product

We need to compute the cross product \(\vec{r} \times \vec{F}\). We can set this up using the determinant of a matrix:

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

### Step 2: Calculate the determinant

To calculate the determinant, we expand it as follows:

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

### Step 3: Calculate each of the 2x2 determinants

1. For \(\hat{i}\):
\[ \begin{vmatrix} -3 & 5 \\ -3 & 5 \end{vmatrix} = (-3)(5) - (-3)(5) = -15 + 15 = 0 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 3 & 5 \\ 2 & 5 \end{vmatrix} = (3)(5) - (2)(5) = 15 - 10 = 5 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 3 & -3 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (2)(-3) = -9 + 6 = -3 \]

### Step 4: Substitute back into the equation

Now substituting these values back into the expression for \(\vec{T}\):

\[ \vec{T} = \hat{i}(0) - \hat{j}(5) + \hat{k}(-3) \]

This simplifies to:

\[ \vec{T} = -5\hat{j} - 3\hat{k} \]

### Final Answer

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

\[ \vec{T} = -5\hat{j} - 3\hat{k} \]

### Conclusion

The correct option is option 3: \(-5\hat{j} - 3\hat{k}\).

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