Question

Let a force $\overrightarrow{F}$ be acting on a body free to rotate about a point $O$ and let $\overrightarrow{r}$ be the position vector of any point $P$ on the line of action of the force. Then torque $\overrightarrow{\tau }$ of this force about point $O$ is defined as :
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
Given, $\overrightarrow{F}=(2\hat{i}+3\hat{j}-\hat{k})N$ and $\overrightarrow{r}=(\hat{i}-\hat{j}+6\hat{k})m$. Find the torque of this force.

• A. $(-17\hat{i}+13\hat{j}+5\hat{k})Nm$
• B. $(17\hat{i}+13\hat{j}+5\hat{k})Nm$
• C. $(-17\hat{i}-13\hat{j}+5\hat{k})Nm$
• D. $(-17\hat{i}+13\hat{j}+10\hat{k})Nm$

Answer 1

The correct option is A $(-17\hat{i}+13\hat{j}+5\hat{k})Nm$

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 6\\ 2& 3& -1\end{array}\right]$
$=\hat{i}(1-18)-\hat{j}(-1-12)+\hat{k}(3+2)$
$=(-17\hat{i}+13\hat{j}+5\hat{k})Nm$