Question

The torque of force $\vec{F}=(2\hat{i}-3\hat{j}+4\hat{k})$ newton acting at a point $\vec{r}=(3\hat{i}+2\hat{j}+3\hat{k})$ metre about origin is:

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

Answer 1

Answer: (C)

Here $:\vec{F}=2\hat{i}-3\hat{j}+4\mathbf{k}N$
Position vector of a point
$\vec{r}=3\hat{i}+2\hat{j}+3\hat{k}m$
The torque acting at a point about the origin is given by
$\vec{\tau }=\vec{r}\times \vec{F}=(3\hat{i}+2\hat{j}+3\hat{k})\times (2\hat{i}-3\hat{j}+4\hat{k})$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 3\\ 2& -3& 4\end{array}\right|$
$=\hat{i}[8-(-9)]-\hat{j}(12-6)+\hat{k}(-9-4)$
$=17\hat{i}-6\hat{j}-13\hat{k}N-m$