Question

A force $\overrightarrow{F}=2\hat{i}+3\hat{j}-\hat{k}$ acts at a point $(2,-3,1)$. Then, magnitude of the torque of this force about point $(0,0,2)$ will be

• A. $6\text{units}$
• B. $3\sqrt{5}\text{units}$
• C. $6\sqrt{5}\text{units}$
• D. None of these

Answer 1

The correct option is C $6\sqrt{5}\text{units}$

As we know,
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
$\overrightarrow{r}=(2-0)\hat{i}+(-3-0)\hat{j}+(1-2)\hat{k}$
$=2\hat{i}-3\hat{j}-\hat{k}$
$\overrightarrow{F}=2\hat{i}+3\hat{j}-\hat{k}$

$\therefore \overrightarrow{\tau }=(2\hat{i}-3\hat{j}-\hat{k})\times (2\hat{i}+3\hat{j}-\hat{k})$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& -1\\ 2& 3& -1\end{array}\right|$
$=\hat{i}(3+3)-\hat{j}(-2+2)+\hat{k}(6+6)$

$\Rightarrow \overrightarrow{\tau }=6\hat{i}+12\hat{k}$
$\left|\overrightarrow{\tau }\right|=\sqrt{6^2+{12}^2}=\sqrt{180}$
$\left|\overrightarrow{\tau }\right|=6\sqrt{5}\text{units}$