Question

A force $\overrightarrow{F}=(\hat{i}+2\hat{j}+3\hat{k})\text{N}$ acts at a point $(4\hat{i}+3\hat{j}-\hat{k})\text{m}$. Then the magnitude of torque about the point $(\hat{i}+2\hat{j}+\hat{k})\text{m}$ will be $\sqrt{x}\text{N-m}$. The value of $x$ is

Answer 1

Given
$\overrightarrow{F}=(\hat{i}+2\hat{j}+3\hat{k})\text{N}$
and, $\overrightarrow{r}=\left[\right(4\hat{i}+3\hat{j}-\hat{k})-(\hat{i}+2\hat{j}+\hat{k}\left)\right]=3\hat{i}+\hat{j}-2\hat{k}$
Torque, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=(3\hat{i}+\hat{j}-2\hat{k})\times (\hat{i}+2\hat{j}+3\hat{k})$
$\overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -2\\ 1& 2& 3\end{array}\right|=7\hat{i}-11\hat{j}+5\hat{k}$
Magnitude of torque, $|\overrightarrow{\tau }|=\sqrt{195}{\text{Nm}}^{-1}$