Question

If force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acts along $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ then the torque acting at that point is

• A. $(14\hat{i} - 38\hat{j} + 16\hat{k})$
• B. $(-14\hat{i} + 34\hat{j} - 16\hat{k})$
• C. $(21\hat{i} + 4\hat{j} + 4\hat{k})$
• D. $(4\hat{i} + 4\hat{j} + 6\hat{k})$

Hint:

For torque: \[ \vec{\tau}=\vec{r}\times \vec{F} \] Be careful with the minus sign in the \(\hat{j}\)-component while expanding the determinant.

Answer 1

The Correct Option is A

Concept:
Torque is given by: \[ \vec{\tau}=\vec{r}\times \vec{F} \] ip

Step 1: Write the vectors.
\[ \vec{r}=7\hat{i}+3\hat{j}+\hat{k} \] \[ \vec{F}=-3\hat{i}+\hat{j}+5\hat{k} \] ip

Step 2: Compute the cross product.
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 7 & 3 & 1 -3 & 1 & 5 \end{vmatrix} \] \[ \vec{\tau} = \hat{i}(3\cdot5-1\cdot1) -\hat{j}(7\cdot5-1\cdot(-3)) +\hat{k}(7\cdot1-3\cdot(-3)) \] \[ \vec{\tau} = 14\hat{i}-38\hat{j}+16\hat{k} \] ip Hence, the correct answer is:
\[ \boxed{(A)\ (14\hat{i}-38\hat{j}+16\hat{k})} \]