The torque of force `vec(F)=-3hat(i)+hat(j)+5hat(k)` acting at the point `vec(r)=7hat(i)+3hat(j)+hat(k)` is ______________?

Options

A. `14hat(i) - 38 hat(j) + hat(k)`
B. `4hat(i) + 4hat(j) + 6hat(k)`
C. `-14hat(i) - 38 hat(j) + 16hat(k)`
D. `-21hat(i) - 3 hat(j) - 5hat(k)`

Answer

To find the torque of the force \(\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}\) acting at the point \(\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}\), we can use the formula for torque, which is given by the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\):

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

### Step 1: Set up the cross product
We can express the vectors \(\vec{r}\) and \(\vec{F}\) as follows:

\[ \vec{r} = 7\hat{i} + 3\hat{j} + 1\hat{k} \]
\[ \vec{F} = -3\hat{i} + 1\hat{j} + 5\hat{k} \]

Now, we can set up the determinant for the cross product:

\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix} \]

### Step 2: Calculate the determinant
To calculate the determinant, we expand it as follows:

\[ \vec{\tau} = \hat{i} \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} 7 & 1 \\ -3 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} 7 & 3 \\ -3 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} = (3)(5) - (1)(1) = 15 - 1 = 14 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 7 & 1 \\ -3 & 5 \end{vmatrix} = (7)(5) - (1)(-3) = 35 + 3 = 38 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 7 & 3 \\ -3 & 1 \end{vmatrix} = (7)(1) - (3)(-3) = 7 + 9 = 16 \]

### Step 3: Combine the results
Now, substituting back into the expression for \(\vec{\tau}\):

\[ \vec{\tau} = 14\hat{i} - 38\hat{j} + 16\hat{k} \]

Thus, the torque vector is:

\[ \vec{\tau} = 14\hat{i} - 38\hat{j} + 16\hat{k} \]

### Final Answer
The torque of the force is:

\[ \vec{\tau} = 14\hat{i} - 38\hat{j} + 16\hat{k} \]

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