Question

The torque of force $\overrightarrow{F}=(2\hat{i}-3\hat{j}+4\hat{k})$ N acting at the point $\overrightarrow{r}=(3\hat{i}+2\hat{j}+3\hat{k})$ m about origin is (in N-m):

• A. $6\hat{i}-6\hat{j}+12\hat{k}$
• B. $17\hat{i}-6\hat{j}-13\hat{k}$
• C. $-6\hat{i}+6\hat{j}-12\hat{k}$
• D. $-17\hat{i}+6\hat{j}+13\hat{k}$

Answer 1

The correct option is B $17\hat{i}-6\hat{j}-13\hat{k}$

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
Thus, $\overrightarrow{\tau }=(3\hat{i}+2\hat{j}+3\hat{k})\times (2\hat{i}-3\hat{j}+4\hat{k})$
Taking the cross product, we get,
$\overrightarrow{\tau }=17\hat{i}-6\hat{j}-13\hat{k}$